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
#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
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
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");
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