Note : All
the programs are tested under Turbo C/C++ compilers.
It is assumed
that,
- Programs run under DOS environment,
- The underlying machine is an x86 system,
- Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based
on this assumptions (for example sizeof(int) == 2 may be assumed).
1. void main()
1. void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler
error: Cannot modify a constant value.
Explanation:
p
is a pointer to a "constant integer". But we tried to change the
value of the "constant integer".
2. main()
{
char
s[ ]="man";
int
i;
for(i=0;s[
i ];i++)
printf("\n%c%c%c%c",s[
i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i],
*(i+s), *(s+i), i[s] are all different ways of expressing the same idea.
Generally array name is the base address
for that array. Here s is the base
address. i is the index
number/displacement from the base address. So, indirecting it with * is same as
s[i]. i[s] may be surprising. But in the
case of C it is same as s[i].
3. main()
{
float
me = 1.1;
double
you = 1.1;
if(me==you)
printf("I
love U");
else
printf("I
hate U");
}
Answer:
I
hate U
Explanation:
For
floating point numbers (float,
double, long double) the values
cannot be predicted exactly. Depending on the number of bytes, the precession
with of the value represented varies.
Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with
less precision than long double.
Rule of Thumb:
Never
compare or at-least be cautious when using floating point numbers with
relational operators (== , >,
<, <=, >=,!= ) .
4. main()
{
static
int var = 5;
printf("%d
",var--);
if(var)
main();
}
Answer:
5
4 3 2 1
Explanation:
When
static storage class is given, it is
initialized once. The change in the value of a static variable is retained even between the function calls. Main
is also treated like any other ordinary function, which can be called
recursively.
5. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf("
%d ",*c);
++q; }
for(j=0;j<5;j++){
printf("
%d ",*p);
++p; }
}
Answer:
2
2 2 2 2 2 3 4 6 5
Explanation:
Initially
pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value
2 will be printed 5 times. In second loop p
itself is incremented. So the values 2 3 4 6 5 will be printed.
6. main()
{
extern
int i;
i=20;
printf("%d",i);
}
Answer:
Linker Error
: Undefined symbol '_i'
Explanation:
extern
storage class in the following declaration,
extern int i;
specifies
to the compiler that the memory for i
is allocated in some other program and that address will be given to the
current program at the time of linking. But linker finds that no other variable
of name i is available in any other
program with memory space allocated for it. Hence a linker error has occurred .
7. main()
{
int
i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d
%d %d %d %d",i,j,k,l,m);
}
Answer:
0
0 1 3 1
Explanation
:
Logical
operations always give a result of 1 or
0 . And also the logical AND (&&) operator has higher priority over
the logical OR (||) operator. So the expression
‘i++ && j++ &&
k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now
the expression is 0 || 2 which evaluates to 1 (because OR operator always gives
1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is
1. The values of other variables are also incremented by 1.
8. main()
{
char
*p;
printf("%d
%d ",sizeof(*p),sizeof(p));
}
Answer:
1
2
Explanation:
The
sizeof() operator gives the number of bytes taken by its operand. P is a
character pointer, which needs one byte for storing its value (a character).
Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the
address of the character pointer sizeof(p) gives 2.
9. main()
{
int
i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer
:
three
Explanation
:
The
default case can be placed anywhere inside the loop. It is executed only when
all other cases doesn't match.
10. main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation
:
-1
is internally represented as all 1's. When left shifted four times the least
significant 4 bits are filled with 0's.The %x format specifier specifies that
the integer value be printed as a hexadecimal value.
11. main()
{
char
string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}
Answer:
Compiler Error
: Type mismatch in redeclaration of function display
Explanation
:
In
third line, when the function display
is encountered, the compiler doesn't know anything about the function display.
It assumes the arguments and return types to be integers, (which is the default
type). When it sees the actual function display,
the arguments and type contradicts with what it has assumed previously. Hence a
compile time error occurs.
12. main()
{
int
c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here
unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However
you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a
constant and not a variable.
13. #define int char
main()
{
int
i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since
the #define replaces the string int by the macro char
14. main()
{
int i=10;
i=!i>14;
printf("i=%d",i);
}
Answer:
i=0
Explanation:
In
the expression !i>14 , NOT (!)
operator has more precedence than ‘ >’ symbol. !
is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).
15. #include<stdio.h>
main()
{
char
s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p +
++*str1-32);
}
Answer:
77
Explanation:
p
is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p
is pointing to '\n' and that is incremented by one." the ASCII value of
'\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1,
str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII
value of 'b' is 98.
Now performing (11 + 98 – 32), we get
77("M");
So we get the output 77 :: "M"
(Ascii is 77).
16. #include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4},
{5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are
trying to access the third 2D(which you are not declared) it will print garbage
values. *q=***a starting address of a is assigned integer pointer. Now q is
pointing to starting address of a. If you print *q, it will print first element
of 3D array.
17. #include<stdio.h>
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler
Error
Explanation:
You
should not initialize variables in declaration
18. #include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct
xx *p;
};
struct yy *q;
};
}
Answer:
Compiler
Error
Explanation:
The
structure yy is nested within structure xx. Hence, the elements are of yy are
to be accessed through the instance of structure xx, which needs an instance of
yy to be known. If the instance is created after defining the structure the
compiler will not know about the instance relative to xx. Hence for nested
structure yy you have to declare member.
19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
20. main()
{
int i=5;
printf("%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The
arguments in a function call are pushed into the stack from left to right. The
evaluation is by popping out from the stack. and the evaluation is from right to left, hence the
result.
Imp Question :
int main()
{
int i=5;
printf("%d%d%d%d%d%d\n",i,i++,i--,++i,--i,i);
printf("1 i : %d\n",i);
printf("2 i++ : %d\n",i++);
printf("3 i-- : %d\n",i--);
printf("4 ++i : %d\n",++i);
printf("5 --i : %d\n",--i);
printf("6 i : %d",i);
return 0;
}
[root@QATS cprog]# ./a.out
545545
1 i : 5
2 i++ : 5
3 i-- : 6
4 ++i : 6
5 --i : 5
6 i : 5
21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
the
macro call square(4) will substituted by 4*4 so the expression becomes i =
64/4*4 . Since / and * has equal priority the expression will be evaluated as
(64/4)*4 i.e. 16*4 = 64
22. main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++
will be parse in the given order
Ø
*p that is value at the location currently
pointed by p will be taken
Ø
++*p the retrieved value will be incremented
Ø
when ; is encountered the location will be
incremented that is p++ will be executed
Hence, in the while loop initial
value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and
pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on.
Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes
“ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print
anything.
23. #include <stdio.h>
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:
50
Explanation:
The
preprocessor directives can be redefined anywhere in the program. So the most
recently assigned value will be taken.
24. #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor
executes as a seperate pass before the execution of the compiler. So textual
replacement of clrscr() to 100 occurs.The input
program to compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100;
is an executable statement but with no action. So it doesn't give any problem
25. main()
{
printf("%p",main);
}
Answer:
Some address will be
printed.
Explanation:
Function
names are just addresses (just like array names are addresses).
main() is also
a function. So the address of function main will be printed. %p in printf
specifies that the argument is an address. They are printed as hexadecimal
numbers.
27) main()
{
clrscr();
}
clrscr();
Answer:
No
output/error
Explanation:
The
first clrscr() occurs inside a function. So it becomes a function call. In the
second clrscr(); is a function declaration (because it is not inside any
function).
28) enum colors {BLACK,BLUE,GREEN}
main()
{
printf("%d..%d..%d",BLACK,BLUE,GREEN);
return(1);
}
Answer:
0..1..2
Explanation:
enum
assigns numbers starting from 0, if not explicitly defined.
29) void main()
{
char far *farther,*farthest;
printf("%d..%d",sizeof(farther),sizeof(farthest));
}
Answer:
4..2
Explanation:
the
second pointer is of char type and not a far pointer
30) main()
{
int i=400,j=300;
printf("%d..%d");
}
Answer:
400..300
Explanation:
printf
takes the values of the first two assignments of the program. Any number of
printf's may be given. All of them take only the first two values. If more
number of assignments given in the program,then printf will take garbage
values.
31) main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}
Answer:
H
Explanation:
*
is a dereference operator & is a reference
operator. They can be applied
any number of times provided it is meaningful. Here p points to
the first character in the string "Hello". *p dereferences it
and so its value is H. Again &
references it to an address and * dereferences it to the value H.
32) main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
Answer:
Compiler
error: Undefined label 'here' in function main
Explanation:
Labels
have functions scope, in other words The scope of the labels is limited to
functions . The label 'here' is available in function fun() Hence it is not
visible in function main.
33) main()
{
static char
names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler
error: Lvalue required in function main
Explanation:
Array
names are pointer constants. So it cannot be modified.
34) void main()
{
int i=5;
printf("%d",i++ + ++i);
}
Answer:
Output
Cannot be predicted exactly.
Explanation:
Side
effects are involved in the evaluation of
i
35) void main()
{
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler
Error
Explanation:
The
expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of
operators.
36) #include<stdio.h>
main()
{
int
i=1,j=2;
switch(i)
{
case 1:
printf("GOOD");
break;
case j:
printf("BAD");
break;
}
}
Answer:
Compiler
Error: Constant expression required in function main.
Explanation:
The
case statement can have only constant expressions (this implies that we cannot
use variable names directly so an error).
Note:
Enumerated
types can be used in case statements.
37) main()
{
int
i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf
returns number of items successfully read and not 1/0. Here 10 is given as input which should have
been scanned successfully. So number of items read is 1.
#define
f(g,g2) g##g2
main()
{
int
var12=100;
printf("%d",f(var,12));
}
Answer:
100
39) main()
{
int
i=0;
for(;i++;printf("%d",i))
;
printf("%d",i);
}
Answer:
1
Explanation:
before
entering into the for loop the checking condition is "evaluated".
Here it evaluates to 0 (false) and comes out of the loop, and i is incremented
(note the semicolon after the for loop).
40) #include<stdio.h>
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
M
Explanation:
p
is pointing to character '\n'.str1 is pointing to character 'a' ++*p
meAnswer:"p is pointing to '\n' and that is incremented by one." the
ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is
11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and
it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is
subtracted from 32.
i.e.
(11+98-32)=77("M");
41) #include<stdio.h>
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct
xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler
Error
Explanation:
Initialization
should not be done for structure members inside the structure declaration
42) #include<stdio.h>
main()
{
struct
xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler
Error
Explanation:
in
the end of nested structure yy a member have to be declared.
43) main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
Answer:
Linker
error: undefined symbol '_i'.
Explanation:
extern
declaration specifies that the variable i is defined somewhere else. The
compiler passes the external variable to be resolved by the linker. So compiler
doesn't find an error. During linking the linker searches for the definition of
i. Since it is not found the linker flags an error.
44) main()
{
printf("%d",
out);
}
int
out=100;
Answer:
Compiler
error: undefined symbol out in function main.
Explanation:
The
rule is that a variable is available for use from the point of declaration.
Even though a is a global variable, it is not available for main. Hence an
error.
45) main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This
is the correct way of writing the previous program.
46) main()
{
show();
}
void
show()
{
printf("I'm the greatest");
}
Answer:
Compier
error: Type mismatch in redeclaration of show.
Explanation:
When
the compiler sees the function show it doesn't know anything about it. So the
default return type (ie, int) is assumed. But when compiler sees the actual
definition of show mismatch occurs since it is declared as void. Hence the
error.
The
solutions are as follows:
1.
declare void show() in main() .
2.
define show() before main().
3.
declare extern void show() before the use of show().
47) main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}
Answer:
100,
100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one.
It can also be viewed as a 1-D array.
2
|
4
|
7
|
8
|
3
|
4
|
2
|
2
|
2
|
3
|
3
|
4
|
100
102 104 106 108
110 112 114
116 118 120
122
thus, for the
first printf statement a, *a, **a give
address of first element . since the
indirection ***a gives the value. Hence, the first line of the output.
for the second
printf a+1 increases in the third dimension thus points to value at 114, *a+1
increments in second dimension thus points to 104, **a +1 increments the first
dimension thus points to 102 and ***a+1 first gets the value at first location
and then increments it by 1. Hence, the output.
48) main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d”
,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(“%d
” ,*p);
p++;
}
}
Answer:
Compiler
error: lvalue required.
Explanation:
Error is in
line with statement a++. The operand must be an lvalue and may be of any of
scalar type for the any operator, array name only when subscripted is an
lvalue. Simply array name is a non-modifiable lvalue.
49) main( )
{
static int
a[ ] = {0,1,2,3,4};
int *p[
] = {a,a+1,a+2,a+3,a+4};
int
**ptr = p;
ptr++;
printf(“\n %d
%d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d
%d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d
%d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d
%d %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
Explanation:
Let us consider the
array and the two pointers with some address
a
0
|
1
|
2
|
3
|
4
|
100 102
104 106 108
p
100
|
102
|
104
|
106
|
108
|
1000
1002 1004 1006
1008
ptr
1000
|
2000
After execution of the instruction ptr++
value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr –
p is value in ptr – starting location of array p, (1002 – 1000) / (scaling
factor) = 1, *ptr – a = value at address
pointed by ptr – starting value of array a, 1002 has a value 102 so the value is (102 – 100)/(scaling factor)
= 1, **ptr is the value stored in the
location pointed by the pointer of ptr =
value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the
output of the firs printf is 1, 1, 1.
After execution of *ptr++ increments value of
the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for
the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of
the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for
the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr
remains the same, the value pointed by the value is incremented by the scaling
factor. So the value in array p at location 1006 changes from 106 10 108,.
Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr –
a = 108 – 100 = 4, **ptr = 4.
50) main( )
{
char
*q;
int j;
for (j=0; j<3; j++) scanf(“%s” ,(q+j));
for (j=0; j<3; j++) printf(“%c” ,*(q+j));
for (j=0; j<3; j++) printf(“%s” ,(q+j));
}
Explanation:
Here we have
only one pointer to type char and since we take input in the same pointer thus
we keep writing over in the same location, each time shifting the pointer value
by 1. Suppose the inputs are MOUSE,
TRACK and VIRTUAL. Then for the first input suppose the pointer starts
at location 100 then the input one is stored as
M |
O |
U |
S
|
E
|
\0
|
When the second
input is given the pointer is incremented as j value becomes 1, so the input is
filled in memory starting from 101.
M |
T |
R |
A |
C |
K |
\0 |
The
third input starts filling from the
location 102
M |
T |
V |
I |
R |
T |
U |
A |
L |
\0
|
This is the final value stored .
The
first printf prints the values at the position q, q+1 and q+2 = M T V
The
second printf prints three strings starting from locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.
51) main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void
pointer is used it can be type casted to any
other type pointer. vp = &ch
stores address of char ch and the next statement prints the value stored
in vp after type casting it to the proper data type pointer. the output is ‘g’.
Similarly the output from second printf
is ‘20’. The third printf statement type casts it to print the string from the
4th value hence the output is ‘fy’.
52) main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char
pointers pointing to start of 4 strings. Then we have ptr which is a pointer to
a pointer of type char and a variable p which is a pointer to a pointer to a
pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next
statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression
is evaluated *++p causes gets value s+1 then the pre decrement is executed and
we get s+1 – 1 = s . the indirection operator now gets the value from the array
of s and adds 3 to the starting address. The string is printed starting from
this position. Thus, the output is ‘ck’.
53) main()
{
int i,
n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i<n; ++i)
{
printf(“%s\n”,x);
x++;
}
}
Answer:
(blank
space)
irl
rl
l
Explanation:
Here a string (a pointer to char) is initialized with
a value “girl”. The strlen function
returns the length of the string, thus n has a value 4. The next statement
assigns value at the nth location (‘\0’) to the first location. Now the string
becomes “\0irl” . Now the printf statement prints the string after each
iteration it increments it starting position.
Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints
nothing and pointer value is incremented. The second time it prints from x[1]
i.e “irl” and the third time it prints “rl” and the last time it prints “l” and
the loop terminates.
54) int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
Answer:
Runtime
error: Abnormal program termination.
assert
failed (i<5), <file name>,<line number>
Explanation:
asserts
are used during debugging to make sure that certain conditions are satisfied.
If assertion fails, the program will terminate reporting the same. After
debugging use,
#undef NDEBUG
and
this will disable all the assertions from the source code. Assertion
is
a good debugging tool to make use of.
55) main()
{
int i=-1;
+i;
printf("i = %d, +i = %d
\n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever
it comes you can just ignore it just because it has no effect in the expressions
(hence the name dummy operator).
56) What are the files which are
automatically opened when a C file is executed?
Answer:
stdin,
stdout, stderr (standard input,standard output,standard error).
57) what will
be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);
Answer
:
a: The SEEK_SET sets the file
position marker to the starting of the file.
b: The SEEK_CUR sets the
file position marker to the current position
of the file.
58) main()
{
char name[10],s[12];
scanf("
\"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First
it checks for the leading white space and discards it.Then it matches with a
quotation mark and then it reads all
character upto another quotation mark.
59) What is the problem with the following
code segment?
while ((fgets(receiving
array,50,file_ptr)) != EOF)
;
Answer
& Explanation:
fgets
returns a pointer. So the correct end of file check is checking for != NULL.
60) main()
{
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main
function calls itself again and again. Each time the function is called its
return address is stored in the call stack. Since there is no condition to
terminate the function call, the call stack overflows at runtime. So it
terminates the program and results in an error.
61) main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler
error (at line number 4): size of v is Unknown.
Explanation:
You
can create a variable of type void * but not of type void, since void is an
empty type. In the second line you are creating variable vptr of type void *
and v of type void hence an error.
62) main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d
%d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
2
5 5
Explanation:
In
first sizeof, str1 is a character pointer so it gives you the size of the
pointer variable. In second sizeof the name str2 indicates the name of the
array whose size is 5 (including the '\0' termination character). The third
sizeof is similar to the second one.
63) main()
{
char not;
not=!2;
printf("%d",not);
}
Answer:
0
Explanation:
!
is a logical operator. In C the value 0 is considered to be the boolean value
FALSE, and any non-zero value is considered to be the boolean value TRUE. Here
2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.
64) #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The
input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor
doesn't replace the values given inside the double quotes. The check by if
condition is boolean value false so it goes to else. In second if -1 is boolean
value true hence "TRUE" is printed.
65) main()
{
int k=1;
printf("%d==1 is
""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1
is TRUE
Explanation:
When
two strings are placed together (or separated by white-space) they are
concatenated (this is called as "stringization" operation). So the
string is as if it is given as "%d==1 is %s". The conditional
operator( ?: ) evaluates to "TRUE".
66) main()
{
int y;
scanf("%d",&y); //
input given is 2000
if( (y%4==0 && y%100 != 0)
|| y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}
Answer:
2000
is a leap year
Explanation:
An
ordinary program to check if leap year or not.
67) #define
max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d
%s",list[0],name);
}
Answer:
Compiler
error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2
is declared of type array of size 5 of characters. So it can be used to declare
the variable name of the type arr2. But it is not the case of arr1. Hence an
error.
Rule of Thumb:
#defines
are used for textual replacement whereas typedefs are used for declaring new
types.
68) int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{'
introduces new block and thus new scope. In the innermost block i is declared
as,
const volatile unsigned
which
is a valid declaration. i is assumed of type int. So printf prints 30. In the
next block, i has value 20 and so printf prints 20. In the outermost block, i
is declared as extern, so no storage space is allocated for it. After
compilation is over the linker resolves it to global variable i (since it is
the only variable visible there). So it prints i's value as 10.
69) main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
Answer:
10
Explanation:
The
variable i is a block level variable and the visibility is inside that block
only. But the lifetime of i is lifetime of the function so it lives upto the
exit of main function. Since the i is still allocated space, *j prints the
value stored in i since j points i.
70) main()
{
int i=-1;
-i;
printf("i = %d, -i = %d
\n",i,-i);
}
Answer:
i
= -1, -i = 1
Explanation:
-i
is executed and this execution doesn't affect the value of i. In printf first
you just print the value of i. After that the value of the expression -i =
-(-1) is printed.
71) #include<stdio.h>
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler
error
Explanation:
i
is a constant. you cannot change the value of constant
72) #include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are
trying to access the third 2D(which you are not declared) it will print garbage
values. *q=***a starting address of a is assigned integer pointer. now q is
pointing to starting address of a.if you print *q meAnswer:it will print first
element of 3D array.
73) #include<stdio.h>
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}
Answer:
hello
5
Explanation:
if
you declare i as register compiler will
treat it as ordinary integer and it will take integer value. i value may
be stored either in register or in memory.
74) main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the
expression i+++j is treated as (i++ + j)
76) struct aaa{
struct
aaa *prev;
int
i;
struct
aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
Above all statements
form a double circular linked list;
abc.next->next->prev->next->i
this
one points to "ghi" node the value of at particular node is 2.
77) struct point
{
int x;
int y;
};
struct
point origin,*pp;
main()
{
pp=&origin;
printf("origin
is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin
is (%d%d)\n",pp->x,pp->y);
}
Answer:
origin
is(0,0)
origin
is(0,0)
Explanation:
pp
is a pointer to structure. we can access the elements of the structure either
with arrow mark or with indirection operator.
Note:
Since
structure point is globally declared x
& y are initialized as zeroes
78) main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int
_l_abc(int i)
{
return(i++);
}
Answer:
9
Explanation:
return(i++)
it will first return i and then increments. i.e. 10 will be returned.
79) main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++
operator when applied to pointers
increments address according to their corresponding data-types.
80) main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler
error
Explanation:
declaration
of convert and format of getc() are wrong.
81) main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int
num1,num2;
{
return num1+num2;
}
Answer:
Compiler
error.
Explanation:
argv[1]
& argv[2] are strings. They are passed to the function sum without
converting it to integer values.
82) # include <stdio.h>
int
one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage
value
Explanation:
ptr
pointer is pointing to out of the array range of one_d.
83) # include<stdio.h>
aaa()
{
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr
is array of pointers to functions of return type int.ptr[0] is assigned to
address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc
respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to
ccc.
85) #include<stdio.h>
main()
{
FILE
*ptr;
char
i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents
of zzz.c followed by an infinite loop
Explanation:
The
condition is checked against EOF, it should be checked against NULL.
86) main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The
value of i is 0. Since this information is enough to determine the truth value
of the boolean expression. So the statement following the if statement is not
executed. The values of i and j remain
unchanged and get printed.
87) main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally
the return value from the function is through the information from the
accumulator. Here _AH is the pseudo global variable denoting the accumulator.
Hence, the value of the accumulator is set 1000 so the function returns value
1000.
88) int i;
main(){
int
t;
for
( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find
the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let
us assume some x= scanf("%d",&i)-t the values during execution
will
be,
t i
x
4 0
-4
3 1
-2
2 2
0
89) main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The
comma operator has associatively from left to right. Only the rightmost value
is returned and the other values are evaluated and ignored. Thus the value of
last variable y is returned to check in if. Since it is a non zero value if
becomes true so, "hello" will be printed.
90) main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c
aptitude");
}
Explanation:
i
is an unsigned integer. It is compared with a signed value. Since the both
types doesn't match, signed is promoted to unsigned value. The unsigned equivalent
of -2 is a huge value so condition becomes false and control comes out of the
loop.
91) In the following pgm add a stmt in the function fun such that the address of
'a'
gets stored in 'j'.
main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The
argument of the function is a pointer to a pointer.
92) What are the following notations of
defining functions known as?
i. int abc(int a,float b)
{
/* some code */
}
ii. int abc(a,b)
int a; float b;
{
/*
some code*/
}
Answer:
i. ANSI C notation
ii.
Kernighan & Ritche notation
93) main()
{
char
*p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The
pointer points to % since it is incremented twice and again decremented by 2,
it points to '%d\n' and 300 is printed.
94)
main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char
a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The
base address is modified only in function and as a result a points to 'b' then
after incrementing to 'c' so bc will be printed.
95) func(a,b)
int
a,b;
{
return( a= (a==b) );
}
main()
{
int
process(),func();
printf("The
value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int
(*pf) ();
int
val1,val2;
{
return((*pf)
(val1,val2));
}
Answer:
The
value if process is 0 !
Explanation:
The
function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers. When this function is
invoked from main, the following substitutions for formal parameters take
place: func for pf, 3 for val1 and 6 for val2. This function returns the result
of the operation performed by the function 'func'. The function func has two
integer parameters. The formal parameters are substituted as 3 for a and 6 for
b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0
which in turn is returned by the function 'process'.
96) void main()
{
static int i=5;
if(--i){
main();
printf("%d
",i);
}
}
Answer:
0 0 0 0
Explanation:
The variable "I" is
declared as static, hence memory for I will be allocated for only once, as it
encounters the statement. The function main() will be called recursively unless
I becomes equal to 0, and since main() is recursively called, so the value of
static I ie., 0 will be printed every time the control is returned.
97) void main()
{
int k=ret(sizeof(float));
printf("\n here value is
%d",++k);
}
int ret(int ret)
{
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The
int ret(int ret), ie., the function name and the argument name can be the same.
Firstly, the function ret() is
called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret
will be 6, as ret is integer hence the value stored in ret will have implicit
type conversion from float to int. The ret is returned in main() it is printed
after and preincrement.
98) void main()
{
char a[]="12345\0";
int i=strlen(a);
printf("here in 3
%d\n",++i);
}
Answer:
here
in 3 6
Explanation:
The char array 'a' will hold the
initialized string, whose length will be counted from 0 till the null
character. Hence the 'I' will hold the value equal to 5, after the
pre-increment in the printf statement, the 6 will be printed.
99) void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u
\n",gotit=--giveit);
}
Answer:
0 65535
Explanation:
100) void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here
\n");
else
printf("Forget
it\n");
}
Answer:
Ok here
Explanation:
Printf will
return how many characters does it print. Hence printing a null character
returns 1 which makes the if statement true, thus "Ok here" is
printed.
void main()
{
void *v;
int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:
Compiler
Error. We cannot apply indirection on type void*.
Explanation:
Void pointer
is a generic pointer type. No pointer arithmetic can be done on it. Void
pointers are normally used for,
1.
Passing generic pointers to functions and returning
such pointers.
2.
As a intermediate pointer type.
3.
Used when the exact pointer type will be known at a
later point of time.
102) void main()
{
int i=i++,j=j++,k=k++;
printf(“%d%d%d”,i,j,k);
}
Answer:
Garbage
values.
Explanation:
An identifier is available to use in program
code from the point of its declaration.
So expressions
such as i = i++ are valid statements.
The i, j and k are automatic variables and so they contain some garbage value. Garbage in is garbage out (GIGO).
103) void main()
{
static int i=i++, j=j++, k=k++;
printf(“i
= %d j = %d k = %d”, i, j, k);
}
Answer:
i
= 1 j = 1 k = 1
Explanation:
Since
static variables are initialized to zero by default.
104) void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage
values
Explanation:
The
inner printf executes first to print some garbage value. The printf returns no
of characters printed and this value also cannot be predicted. Still the outer
printf prints something and so returns a
non-zero value. So it encounters the break statement and comes out of the while
statement.
104) main()
{
unsigned int i=10;
while(i-->=0)
printf("%u
",i);
}
Answer:
10
9 8 7 6 5 4 3 2 1 0 65535 65534…..
Explanation:
Since i is an
unsigned integer it can never become negative. So the expression i--
>=0 will always be true, leading to
an infinite loop.
105) #include<conio.h>
main()
{
int x,y=2,z,a;
if(x=y%2) z=2;
a=2;
printf("%d %d ",z,x);
}
Answer:
Garbage-value
0
Explanation:
The
value of y%2 is 0. This value is assigned to x. The condition reduces to if (x)
or in other words if(0) and so z goes uninitialized.
Thumb
Rule:
Check all control paths to write bug free code.
106) main()
{
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a cancels out. The result
is as simple as 1 + 3 = 4 !
107) #define prod(a,b) a*b
main()
{
int x=3,y=4;
printf("%d",prod(x+2,y-1));
}
Answer:
10
Explanation:
The macro expands and evaluates to
as:
x+2*y-1 => x+(2*y)-1 => 10
108) main()
{
unsigned int i=65000;
while(i++!=0);
printf("%d",i);
}
Answer:
1
Explanation:
Note the
semicolon after the while statement. When the value of i becomes 0 it comes out
of while loop. Due to post-increment on i the value of i while printing is 1.
109) main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
Explanation:
Unary + is the only dummy operator in C.
So it has no effect on the expression and now the while loop is, while(i--!=0) which is false and so
breaks out of while loop. The value –1 is printed due to the post-decrement
operator.
113) main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%lf\n",f%g);
printf("%lf\n",fmod(f,g));
}
Answer:
Line
no 5: Error: Lvalue required
Line
no 6: Cannot apply leftshift to float
Line
no 7: Cannot apply mod to float
Explanation:
Enumeration
constants cannot be modified, so you cannot apply ++.
Bit-wise
operators and % operators cannot be applied on float values.
fmod()
is to find the modulus values for floats as % operator is for ints.
110) main()
{
int i=10;
f(i++,i++,i++);
printf(" %d",i);
}
void pascal
f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}
Answer:
Compiler
error: unknown type integer
Compiler
error: undeclared function write
Explanation:
Pascal
keyword doesn’t mean that pascal code can be used. It means that the function
follows Pascal argument passing mechanism in calling the functions.
111)
void pascal f(int i,int j,int k)
{
printf(“%d
%d %d”,i, j, k);
}
void cdecl f(int
i,int j,int k)
{
printf(“%d
%d %d”,i, j, k);
}
main()
{
int i=10;
printf(" %d\n",i);
i=10;
printf("
%d",i);
}
Answer:
10
11 12 13
12
11 10 13
Explanation:
Pascal
argument passing mechanism forces the arguments to be called from left to
right. cdecl is the normal C argument passing mechanism where the arguments are
passed from right to left.
112).
What is the output of the program given below
main()
{
signed char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
-128
Explanation
Notice the
semicolon at the end of the for loop. THe initial value of the i is set to 0.
The inner loop executes to increment the value from 0 to 127 (the positive
range of char) and then it rotates to the negative value of -128. The condition
in the for loop fails and so comes out of the for loop. It prints the current
value of i that is -128.
113) main()
{
unsigned char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
infinite loop
Explanation
The
difference between the previous question and this one is that the char is
declared to be unsigned. So the i++ can never yield negative value and i>=0
never becomes false so that it can come out of the for loop.
114) main()
{
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer:
Behavior is
implementation dependent.
Explanation:
The detail if the char is signed/unsigned by default is implementation
dependent. If the implementation treats the char to be signed by default the
program will print –128 and terminate. On the other hand if it considers char
to be unsigned by default, it goes to infinite loop.
Rule:
You can write programs that have implementation dependent behavior. But
dont write programs that depend on such behavior.
115) Is the
following statement a declaration/definition. Find what does it mean?
int
(*x)[10];
Answer
Definition.
x is a pointer to array of(size 10)
integers.
Apply clock-wise rule to
find the meaning of this definition.
116). What is
the output for the program given below
typedef enum errorType{warning, error,
exception,}error;
main()
{
error g1;
g1=1;
printf("%d",g1);
}
Answer
Compiler error: Multiple
declaration for error
Explanation
The name error is used in the two meanings. One means that it is a
enumerator constant with value 1. The another use is that it is a type name
(due to typedef) for enum errorType. Given a situation the compiler cannot
distinguish the meaning of error to know in what sense the error is used:
error g1;
g1=error;
// which error it refers in each
case?
When the compiler can distinguish between usages then it will not issue
error (in pure technical terms, names can only be overloaded in different
namespaces).
Note: the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an
error. An extra comma is valid and is provided just for programmer’s
convenience.
117) typedef struct error{int warning, error,
exception;}error;
main()
{
error g1;
g1.error =1;
printf("%d",g1.error);
}
Answer
1
Explanation
The
three usages of name errors can be distinguishable by the compiler at any
instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can
be used only by preceding the error by struct kayword as in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used
only after . (dot) or -> (arrow) operator preceded by the variable name as
in :
g1.error =1;
printf("%d",g1.error);
typedef
struct error{int warning, error, exception;}error;
This can be used
to define variables without using the preceding struct keyword as in:
error g1;
Since the compiler
can perfectly distinguish between these three usages, it is perfectly legal and
valid.
Note
This
code is given here to just explain the concept behind. In real programming
don’t use such overloading of names. It reduces the readability of the code. Possible
doesn’t mean that we should use it!
118) #ifdef something
int
some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
Compiler error :
undefined symbol some
Explanation:
This is a very
simple example for conditional compilation. The name something is not already
known to the compiler making the declaration
int some = 0;
effectively removed from the source code.
119) #if something == 0
int
some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer
0 0
Explanation
This code is to
show that preprocessor expressions are not the same as the ordinary
expressions. If a name is not known the preprocessor treats it to be equal to
zero.
120. What is the
output for the following program
main()
{
int arr2D[3][3];
printf("%d\n", ((arr2D==*
arr2D)&&(* arr2D == arr2D[0])) );
}
Answer
1
Explanation
This is due to the close relation between the arrays and pointers. N dimensional arrays are made up of (N-1) dimensional arrays. arr2D is made up of a 3 single arrays that contains 3 integers each .
The name arr2D refers to the beginning of all the 3 arrays. *arr2D refers to the start of the first 1D array (of 3 integers) that is the same address as arr2D. So the expression (arr2D == *arr2D) is true (1).
This is due to the close relation between the arrays and pointers. N dimensional arrays are made up of (N-1) dimensional arrays. arr2D is made up of a 3 single arrays that contains 3 integers each .
The name arr2D refers to the beginning of all the 3 arrays. *arr2D refers to the start of the first 1D array (of 3 integers) that is the same address as arr2D. So the expression (arr2D == *arr2D) is true (1).
Since both
parts of the expression evaluates to true the result is true(1) and the same is
printed.
121) void main()
{
if(~0
== (unsigned int)-1)
printf(“You can
answer this if you know how values are represented in memory”);
}
Answer
You can answer this if you know how values are represented in memory
Explanation
~ (tilde
operator or bit-wise negation operator) operates on 0 to produce all ones to
fill the space for an integer. –1 is represented in unsigned value as all 1’s
and so both are equal.
122) int
swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
printf("x= %d y =
%d\n",x,y);
}
Answer
x = 20 y = 10
Explanation
This is one way
of swapping two values. Simple checking will help understand this.
123) main()
{
char
*p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b
124) main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue
required in function main
Explanation:
++i yields an
rvalue. For postfix ++ to operate an
lvalue is required.
125) main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:
b
Explanation:
There is no
difference between the expression ++*(p++) and ++*p++. Parenthesis just works
as a visual clue for the reader to see which expression is first evaluated.
126)
int
aaa() {printf(“Hi”);}
int
bbb(){printf(“hello”);}
iny
ccc(){printf(“bye”);}
main()
{
int
( * ptr[3]) ();
ptr[0]
= aaa;
ptr[1]
= bbb;
ptr[2]
=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
int (*
ptr[3])() says that ptr is an array of pointers to functions that takes no
arguments and returns the type int. By the assignment ptr[0] = aaa; it means
that the first function pointer in the array is initialized with the address of
the function aaa. Similarly, the other two array elements also get initialized
with the addresses of the functions bbb and ccc. Since ptr[2] contains the
address of the function ccc, the call to the function ptr[2]() is same as
calling ccc(). So it results in printing
"bye".
127)
main()
{
int i=5;
printf(“%d”,i=++i
==6);
}
Answer:
1
Explanation:
The expression
can be treated as i = (++i==6), because == is of higher precedence than =
operator. In the inner expression, ++i is equal to 6 yielding true(1). Hence
the result.
128) main()
{
char
p[ ]="%d\n";
p[1]
= 'c';
printf(p,65);
}
Answer:
A
Explanation:
Due to the
assignment p[1] = ‘c’ the string becomes, “%c\n”. Since this string becomes the
format string for printf and ASCII value of 65 is ‘A’, the same gets printed.
129) void ( * abc( int, void ( *def) () ) ) ();
Answer::
abc is a
ptr to a function which takes 2
parameters .(a). an integer variable.(b).
a ptrto a funtion which returns void. the return type of the function
is void.
Explanation:
Apply
the clock-wise rule to find the result.
130) main()
{
while
(strcmp(“some”,”some\0”))
printf(“Strings
are not equal\n”);
}
Answer:
No
output
Explanation:
Ending the
string constant with \0 explicitly makes no difference. So “some” and “some\0”
are equivalent. So, strcmp returns 0 (false) hence breaking out of the while
loop.
131) main()
{
char
str1[] = {‘s’,’o’,’m’,’e’};
char
str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while
(strcmp(str1,str2))
printf(“Strings
are not equal\n”);
}
Answer:
“Strings
are not equal”
“Strings
are not equal”
….
Explanation:
If a string
constant is initialized explicitly with characters, ‘\0’ is not appended
automatically to the string. Since str1 doesn’t have null termination, it
treats whatever the values that are in the following positions as part of the
string until it randomly reaches a ‘\0’. So str1 and str2 are not the same,
hence the result.
132) main()
{
int
i = 3;
for
(;i++=0;) printf(“%d”,i);
}
Answer:
Compiler
Error: Lvalue required.
Explanation:
As we know that
increment operators return rvalues and
hence it cannot appear on the left hand side of an assignment operation.
133) void main()
{
int
*mptr, *cptr;
mptr
= (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int
*cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value
0
Explanation:
The memory
space allocated by malloc is uninitialized, whereas calloc returns the
allocated memory space initialized to zeros.
134) void main()
{
static
int i;
while(i<=10)
(i>2)?i++:i--;
printf(“%d”, i);
}
Answer:
32767
Explanation:
Since i is
static it is initialized to 0. Inside the while loop the conditional operator
evaluates to false, executing i--. This continues till the integer value
rotates to positive value (32767). The while condition becomes false and hence,
comes out of the while loop, printing the i value.
135) main()
{
int
i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d
%d",i,j);
}
Answer:
10
10
Explanation:
The
Ternary operator ( ? : ) is equivalent for if-then-else statement. So the question
can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j
= j;
136) 1. const char *a;
2. char* const
a;
3. char const
*a;
-Differentiate
the above declarations.
Answer:
1. 'const'
applies to char * rather than 'a' ( pointer to a constant char )
*a='F' : illegal
a="Hi" : legal
2.
'const' applies to 'a' rather than to
the value of a (constant pointer to char )
*a='F' : legal
a="Hi" : illegal
3.
Same as 1.
137) main()
{
int
i=5,j=10;
i=i&=j&&10;
printf("%d
%d",i,j);
}
Answer:
1
10
Explanation:
The expression
can be written as i=(i&=(j&&10)); The inner expression
(j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence
the result.
138) main()
{
int
i=4,j=7;
j = j || i++ &&
printf("YOU CAN");
printf("%d
%d", i, j);
}
Answer:
4
1
Explanation:
The boolean expression needs to be evaluated
only till the truth value of the expression is not known. j is not equal to
zero itself means that the expression’s truth value is 1. Because it is
followed by || and true || (anything)
=> true where (anything) will not be evaluated. So the remaining
expression is not evaluated and so the value of i remains the same.
Similarly when
&& operator is involved in an expression, when any of the operands
become false, the whole expression’s truth value becomes false and hence the
remaining expression will not be evaluated.
false
&& (anything) => false where (anything) will not be evaluated.
139) main()
{
register
int a=2;
printf("Address of a =
%d",&a);
printf("Value
of a = %d",a);
}
Answer:
Compier
Error: '&' on register variable
Rule
to Remember:
& (address of ) operator cannot be applied
on register variables.
140) main()
{
float
i=1.5;
switch(i)
{
case
1: printf("1");
case
2: printf("2");
default
: printf("0");
}
}
Answer:
Compiler
Error: switch expression not integral
Explanation:
Switch statements can be applied only to
integral types.
141) main()
{
extern
i;
printf("%d\n",i);
{
int i=20;
printf("%d\n",i);
}
}
Answer:
Linker
Error : Unresolved external symbol i
Explanation:
The identifier
i is available in the inner block and so using extern has no use in resolving
it.
142) main()
{
int
a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d %d
%d",a,*f1,*f2);
}
Answer:
16
16 16
Explanation:
f1 and f2 both
refer to the same memory location a. So changes through f1 and f2 ultimately
affects only the value of a.
143) main()
{
char
*p="GOOD";
char a[ ]="GOOD";
printf("\n
sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p), sizeof(*p),
strlen(p));
printf("\n sizeof(a) = %d,
strlen(a) = %d", sizeof(a), strlen(a));
}
Answer:
sizeof(p)
= 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a) = 5, strlen(a) = 4
Explanation:
sizeof(p)
=> sizeof(char*) => 2
sizeof(*p) => sizeof(char) =>
1
Similarly,
sizeof(a) => size of the
character array => 5
When sizeof operator is applied to an array it returns the sizeof the
array and
it is not the same as the sizeof the pointer variable. Here the sizeof(a) where
a is the character array and the size of the array is 5 because the space
necessary for the terminating NULL character should also be taken into account.
144) #define
DIM( array, type) sizeof(array)/sizeof(type)
main()
{
int arr[10];
printf(“The
dimension of the array is %d”, DIM(arr, int));
}
Answer:
10
Explanation:
The size of integer array of 10 elements is 10 *
sizeof(int). The macro expands to sizeof(arr)/sizeof(int) => 10 *
sizeof(int) / sizeof(int) => 10.
145) int
DIM(int array[])
{
return
sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf(“The
dimension of the array is %d”, DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as
arguments and only the pointers can be passed. So the argument is
equivalent to int * array (this is one of the very few places where [] and *
usage are equivalent). The return statement becomes, sizeof(int *)/ sizeof(int)
that happens to be equal in this case.
146) main()
{
static int a[3][3]={1,2,3,4,5,6,7,8,9};
int i,j;
static *p[]={a,a+1,a+2};
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j),
*(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
}
}
Answer:
1 1
1 1
2 4
2 4
3 7
3 7
4 2
4 2
5 5
5 5
6 8
6 8
7 3
7 3
8 6
8 6
9 9
9 9
Explanation:
*(*(p+i)+j)
is equivalent to p[i][j].
147) main()
{
void
swap();
int x=10,y=8;
swap(&x,&y);
printf("x=%d y=%d",x,y);
}
void swap(int
*a, int *b)
{
*a ^= *b,
*b ^= *a, *a ^= *b;
}
Answer:
x=10
y=8
Explanation:
Using ^ like
this is a way to swap two variables without using a temporary variable and that
too in a single statement.
Inside main(),
void swap(); means that swap is a function that may take any number of
arguments (not no arguments) and returns nothing. So this doesn’t issue a
compiler error by the call swap(&x,&y); that has two arguments.
This convention
is historically due to pre-ANSI style (referred to as Kernighan and Ritchie
style) style of function declaration. In that style, the swap function will be
defined as follows,
void swap()
int *a, int *b
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
where the
arguments follow the (). So naturally the declaration for swap will look like,
void swap() which means the swap can take any number of arguments.
148) main()
{
int
i = 257;
int *iPtr = &i;
printf("%d
%d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
1
1
Explanation:
The integer
value 257 is stored in the memory as, 00000001 00000001, so the individual
bytes are taken by casting it to char * and get printed.
149) main()
{
int
i = 258;
int *iPtr = &i;
printf("%d
%d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
2
1
Explanation:
The integer
value 257 can be represented in binary as, 00000001 00000001. Remember that the
INTEL machines are ‘small-endian’ machines. Small-endian
means that the lower order bytes are stored in the higher memory addresses and
the higher order bytes are stored in lower addresses. The integer value 258
is stored in memory as: 00000001 00000010.
150) main()
{
int
i=300;
char *ptr = &i;
*++ptr=2;
printf("%d",i);
}
Answer:
556
Explanation:
The integer
value 300 in binary notation is:
00000001 00101100. It is stored in
memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr =
2 makes the memory representation as: 00101100 00000010. So the integer
corresponding to it is 00000010 00101100 => 556.
151) #include <stdio.h>
main()
{
char
* str = "hello";
char
* ptr = str;
char
least = 127;
while
(*ptr++)
least = (*ptr<least )
?*ptr :least;
printf("%d",least);
}
Answer:
0
Explanation:
After ‘ptr’ reaches
the end of the string the value pointed by ‘str’ is ‘\0’. So the value of ‘str’
is less than that of ‘least’. So the value of ‘least’ finally is 0.
152) Declare an array of N pointers to functions
returning pointers to functions returning pointers to characters?
Answer:
(char*(*)(
)) (*ptr[N])( );
153) main()
{
struct
student
{
char name[30];
struct date dob;
}stud;
struct
date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno,
&student.dob.day, &student.dob.month, &student.dob.year);
}
Answer:
Compiler
Error: Undefined structure date
Explanation:
Inside the
struct definition of ‘student’ the member of type struct date is given. The
compiler doesn’t have the definition of date structure (forward reference is not allowed in C in this case)
so it issues an error.
154) main()
{
struct
date;
struct
student
{
char
name[30];
struct
date dob;
}stud;
struct
date
{
int day,month,year;
};
scanf("%s%d%d%d",
stud.rollno, &student.dob.day, &student.dob.month,
&student.dob.year);
}
Answer:
Compiler
Error: Undefined structure date
Explanation:
Only
declaration of struct date is available inside the structure definition of
‘student’ but to have a variable of type struct date the definition of the
structure is required.
155) There were 10 records stored in
“somefile.dat” but the following program printed 11 names. What went wrong?
void main()
{
struct
student
{
char
name[30], rollno[6];
}stud;
FILE
*fp = fopen(“somefile.dat”,”r”);
while(!feof(fp))
{
fread(&stud,
sizeof(stud), 1 , fp);
puts(stud.name);
}
}
Explanation:
fread reads 10
records and prints the names successfully. It will return EOF only when fread
tries to read another record and fails reading EOF (and returning EOF). So it
prints the last record again. After this only the condition feof(fp) becomes
false, hence comes out of the while loop.
156) Is there any difference between the two declarations,
1.
int foo(int *arr[]) and
2.
int foo(int *arr[2])
Answer:
No
Explanation:
Functions can only
pass pointers and not arrays. The numbers that are allowed inside the [] is
just for more readability. So there is no difference between the two
declarations.
157) What is the subtle error in the following code segment?
void fun(int n,
int arr[])
{
int
*p=0;
int
i=0;
while(i++<n)
p = &arr[i];
*p = 0;
}
Answer
& Explanation:
If the body of
the loop never executes p is assigned no address. So p remains NULL where *p =0
may result in problem (may rise to runtime error “NULL pointer assignment” and
terminate the program).
158) What is wrong with the following code?
int *foo()
{
int
*s = malloc(sizeof(int)100);
assert(s
!= NULL);
return
s;
}
Answer
& Explanation:
assert macro
should be used for debugging and finding out bugs. The check s != NULL is for
error/exception handling and for that assert shouldn’t be used. A plain if and
the corresponding remedy statement has to be given.
159) What is the hidden bug with the following statement?
assert(val++
!= 0);
Answer
& Explanation:
Assert macro is
used for debugging and removed in release version. In assert, the experssion
involves side-effects. So the behavior of the code becomes different in case of
debug version and the release version thus leading to a subtle bug.
Rule
to Remember:
Don’t use expressions that have side-effects
in assert statements.
160) void main()
{
int *i =
0x400; // i points to the address 400
*i = 0; // set the value of memory
location pointed by i;
}
Answer:
Undefined
behavior
Explanation:
The second
statement results in undefined behavior because it points to some location
whose value may not be available for modification. This
type of pointer in which the non-availability of the implementation of the
referenced location is known as 'incomplete type'.
161) #define assert(cond) if(!(cond)) \
(fprintf(stderr, "assertion failed: %s,
file %s, line %d \n",#cond,\
__FILE__,__LINE__), abort())
void main()
{
int i = 10;
if(i==0)
assert(i < 100);
else
printf("This statement becomes else
for if in assert macro");
}
Answer:
No
output
Explanation:
The else part in
which the printf is there becomes the else for if in the assert macro. Hence
nothing is printed.
The solution is
to use conditional operator instead of if statement,
#define
assert(cond) ((cond)?(0): (fprintf (stderr, "assertion failed: \ %s, file
%s, line %d \n",#cond, __FILE__,__LINE__), abort()))
Note:
However this
problem of “matching with nearest else” cannot be solved by the usual method of
placing the if statement inside a block like this,
#define
assert(cond) { \
if(!(cond))
\
(fprintf(stderr, "assertion failed: %s,
file %s, line %d \n",#cond,\
__FILE__,__LINE__), abort()) \
}
162) Is the following code legal?
struct a
{
int
x;
struct a b;
}
Answer:
No
Explanation:
Is
it not legal for a structure to contain a member that is of the same
type as in this
case. Because this will cause the structure declaration to be recursive without
end.
163) Is the following code legal?
struct a
{
int
x;
struct a *b;
}
Answer:
Yes.
Explanation:
*b is a pointer
to type struct a and so is legal. The compiler knows, the size of the pointer
to a structure even before the size of the structure
is
determined(as you know the pointer to any type is of same size). This type of
structures is known as ‘self-referencing’ structure.
164) Is the following code legal?
typedef struct a
{
int
x;
aType *b;
}aType
Answer:
No
Explanation:
The typename
aType is not known at the point of declaring the structure (forward references
are not made for typedefs).
165) Is the following code legal?
typedef struct a
aType;
struct a
{
int
x;
aType
*b;
};
Answer:
Yes
Explanation:
The typename
aType is known at the point of declaring the structure, because it is already
typedefined.
166) Is the following code legal?
void main()
{
typedef
struct a aType;
aType
someVariable;
struct
a
{
int
x;
aType *b;
};
}
Answer:
No
Explanation:
When
the declaration,
typedef
struct a aType;
is encountered
body of struct a is not known. This is known as ‘incomplete types’.
167) void main()
{
printf(“sizeof
(void *) = %d \n“, sizeof( void *));
printf(“sizeof
(int *) = %d \n”, sizeof(int *));
printf(“sizeof
(double *) = %d \n”, sizeof(double *));
printf(“sizeof(struct
unknown *) = %d \n”, sizeof(struct unknown *));
}
Answer :
sizeof
(void *) = 2
sizeof
(int *) = 2
sizeof
(double *) = 2
sizeof(struct
unknown *) = 2
Explanation:
The
pointer to any type is of same size.
168) char inputString[100] = {0};
To get string
input from the keyboard which one of the following is better?
1) gets(inputString)
2) fgets(inputString,
sizeof(inputString), fp)
Answer
& Explanation:
The second one
is better because gets(inputString) doesn't know the size of the string passed
and so, if a very big input (here, more than 100 chars) the charactes will be
written past the input string. When fgets is used with stdin performs the same
operation as gets but is safe.
169) Which version do you prefer of the following two,
1)
printf(“%s”,str); // or the more curt
one
2)
printf(str);
Answer
& Explanation:
Prefer the
first one. If the str contains any
format characters like %d then it will result in a subtle bug.
170) void main()
{
int
i=10, j=2;
int
*ip= &i, *jp = &j;
int
k = *ip/*jp;
printf(“%d”,k);
}
Answer:
Compiler
Error: “Unexpected end of file in comment started in line 5”.
Explanation:
The programmer
intended to divide two integers, but by the “maximum munch” rule, the compiler
treats the operator sequence / and * as /* which happens to be the starting of
comment. To force what is intended by the programmer,
int k = *ip/ *jp;
// give space explicity separating / and *
//or
int k = *ip/(*jp);
// put braces to force the intention
will solve the problem.
171) void main()
{
char ch;
for(ch=0;ch<=127;ch++)
printf(“%c %d \n“, ch, ch);
}
Answer:
Implementaion dependent
Explanation:
The char type
may be signed or unsigned by default. If it is signed then ch++ is executed
after ch reaches 127 and rotates back to -128. Thus ch is always smaller than 127.
172) Is this code legal?
int
*ptr;
ptr = (int *)
0x400;
Answer:
Yes
Explanation:
The
pointer ptr will point at the integer in the memory location 0x400.
173) main()
{
char
a[4]="HELLO";
printf("%s",a);
}
Answer:
Compiler
error: Too many initializers
Explanation:
The array a is
of size 4 but the string constant requires 6 bytes to get stored.
174) main()
{
char
a[4]="HELL";
printf("%s",a);
}
Answer:
HELL%@!~@!@???@~~!
Explanation:
The character
array has the memory just enough to hold the string “HELL” and doesnt have
enough space to store the terminating null character. So it prints the HELL
correctly and continues to print garbage values till it accidentally comes across a NULL character.
175) main()
{
int
a=10,*j;
void *k;
j=k=&a;
j++;
k++;
printf("\n
%u %u ",j,k);
}
Answer:
Compiler
error: Cannot increment a void pointer
Explanation:
Void pointers
are generic pointers and they can be used only when the type is not known and
as an intermediate address storage type. No pointer arithmetic can be done on
it and you cannot apply indirection operator (*) on void pointers.
176) main()
{
extern
int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
int
i;
177) Printf can be implemented by using __________ list.
Answer:
Variable
length argument lists
178) char
*someFun()
{
char *temp = “string constant";
return temp;
}
int main()
{
puts(someFun());
}
Answer:
string constant
Explanation:
The program suffers no problem and
gives the output correctly because the character constants are stored in
code/data area and not allocated in stack, so this doesn’t lead to dangling
pointers.
179) char *someFun1()
{
char temp[ ] = “string";
return temp;
}
char *someFun2()
{
char temp[ ] = {‘s’,
‘t’,’r’,’i’,’n’,’g’};
return temp;
}
int main()
{
puts(someFun1());
puts(someFun2());
}
Answer:
Garbage
values.
Explanation:
Both
the functions suffer from the problem of dangling pointers. In someFun1() temp
is a character array and so the space for it is allocated in heap and is
initialized with character string “string”. This is created dynamically as the
function is called, so is also deleted dynamically on exiting the function so
the string data is not available in the calling function main() leading to
print some garbage values. The function someFun2() also suffers from the same
problem but the problem can be easily identified in this case.
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