61. main()
{
41printf("%p",main);
}8
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.
62. 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).
63. 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.
64. 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
65. 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.
66. 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.
67. 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.
68. 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.
69. 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
70. 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.
71. #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.
72. 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.
73. #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer : 100
74. 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).
75. #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 "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 "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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{
41printf("%p",main);
}8
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.
62. 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).
63. 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.
64. 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
65. 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.
66. 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.
67. 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.
68. 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.
69. 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
70. 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.
71. #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.
72. 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.
73. #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer : 100
74. 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).
75. #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 "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 "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");