when the function f returns a value then why is it not getting printed ?

Question

#include<stdio.h>

int f();

int main(){
    printf("hello");
    int a=f();
    printf("%d",a);
    return 0;
}

int f(){
    main();
    return 8;
}

Why is 8 not printed here? I am a bit confused that after f returns, it returns 8, so a must be assigned 8. Then how come it is not getting printed?

Comments

it will fall in infinite loop becoz in main(), you are calling f() and in f(), you are calling main().

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 yeah that's right. @vivek

Answer 1

Because the print statement will never be executed. Until the stack overflow there will be call from main to function f and vice-versa.

Comments

But then how is f terminating ?

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f will not terminate Until the stack overflows.

Answer 2

I have executed it in ideone nd it is showing infinite no of hello nd it should be bccause in the first line of the main function 

printf( "hello") will print a single hello.In the next line we r calling f() which is calling main() which first prints hello and then again calls f() and then main().............it goes on infinetly .....so the programm is not going to execute  

 printf("%d",a);
    return 0;

of main function ever and return 8 of f() function .so,a will never contain 8 nd 8 will never be printed .

calling main recursively should be avoided but using main recursively is not illegal. See the link 

http://stackoverflow.com/questions/21941040/calling-main-function-in-c