what is the output

Question

#include <stdio.h>
int main(){
    int k = m();
    printf("%d ",k);
}
void m(){
    printf("hello\n");
}

https://gateoverflow.in/?qa=blob&qa_blobid=15997855123266011837

i think code should throws compiler error because no declaration of m() before calling it.so compiler assume return type must be int but is void.but code compile and run correctly.please explain

Comments

Modern compiler don,t give error in this case they simply warn by giving the following msg,

conflicting types for 'm'

previous implicit declaration of 'm' was here

This is happening bcz of implicit declaration is int for m.So simply change the called function to 'int' .No warning .

Answer 1

In C99 strict, we can not do this. In C89/C90 or gcc 4.8.4+, we need not need to pre-declare (or prototype) a function; it would be implicitly declared as function taking undefined list of arguments and having return type 'int'.
Although type conpflict 'warning' would occur.

Below the results using gcc 4.8.4

In of case c99 strict : compilation gives errors:

prog.c: In function 'main':
prog.c:3:10: error: implicit declaration of function 'm' [-Werror=implicit-function-declaration]
  int k = m();
          ^
prog.c: At top level:
prog.c:6:6: error: conflicting types for 'm' [-Werror]
 void m(){
      ^
prog.c:3:10: note: previous implicit declaration of 'm' was here
  int k = m();
          ^
cc1: all warnings being treated as errors

Answer 2

It will output 6