C for loop Tricky Interview Questions and Answers

#include<stdio.h>
int main()
{
	static int i;
	for(i++;++i;i++)
	{
		printf("%d ", i);
		if(i == 6) 
		break;
	}
	return 0;
}

x

 

Option: C

Explanation

By Default, the values of static int variable is set to zero by C compiler. i.e) i = 0;
Clearly, the for loop in this program is infinite for loop, ie) At first iteration the program looks like for( 0;2;2 ) that is it 2 4 6 8 10 .. and so on. By this program we breaks when i = 6.

#include<stdio.h>
int fun();
int main(){
	for(fun();fun();fun()) 
	{
	printf("%d ", fun());	
	}
	return 0;
}
int fun()
{
    int static num = 10;
    return num--;
}

x

 

Option: A

Explanation

Clearly for(fun();fun();fun())

At First Execution

for(10;9;fun()) //condition is true
printf("%d", 8); //8 prints

At Second Execution

for(10;fun();7)
for(7;6;fun())
printf("%d", 5); //5 prints

At Third Execution

for(7;fun();4)
for(4;3;fun())
printf("%d", 2); //2 prints

At fourth Execution

for(4;fun();1)
for(1;0;fun())//condition fails and program terminated
Thus outputted 8 5 2

#include<stdio.h>
int main()
{
	for(;;) 
	{
		printf("%d ", 10);
	}
	return 0;
}

x

 

Option: C

Explanation

Yes, it is infinite for loop.

#include<stdio.h>
int main()
{
	int fun = {
	printf("C for loop ")
	};
	int x = 5;
	for(x=0;x<=fun;x++)
	{
	printf("%x ", x);
	}
	return 0;
}

x

 

Option: B

Explanation

for loop executes first.
for(x=0;x<=fun;x++)
ie) for(x=0; x=printf("C for loop"); x++)
Now printf in conditional place of for loop prints C for loop and the count in that string is replace in conditional place of for loop
ie) for(x=0; x<=10;x++)
thus outputted C for loop 0 1 2 3 4 5 6 7 8 9 a
here a represent 10 in hexadecimal.

#include<stdio.h>
int main()
{
	int i;
	for(i = 0;i<=3;i++);
	printf("%d", i);
	return 0;
}

x

 

Option: C

Explanation

Although for loop terminated within itself, it iterates for 4 times and 4 alone is outputted.