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.
38) #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.
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