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Function with no return type specified in C

I came across this piece of code in C:

#include <stdio.h>
main( )
{
 int i = 5;
 workover(i);
 printf("%d",i);
}
workover(i)
int i;
{
 i = i*i;
 return(i);
}

I want to know how is it allowed to declare a function "overhaul"? What happens when we don't mention the return type of a function? (can we return anything?). The parameter is also just a variable name, how does it work?

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3 answers


If you don't specify a return type or a parameter type, C will implicitly declare it as int

.

This is a "feature" from earlier versions of C (C89 and C90) but is currently considered bad practice. Since the C99 (1999) standard no longer allows this, a compiler targeting C99 or later will likely give you a warning similar to the following:



program.c: At top level:
program.c:8:1: warning: return type defaults to ‘int’
 workover(i)
 ^
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The function declaration syntax was used in older versions of C and is still valid, so the "workover (i) int i;" equivalent to "overhaul (int i)". Although, I think it can still generate warnings or even errors, depending on which compiler options you use.



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When I compile your code as $ gcc common.c -o common.exe -Wall

(Trying it through Cygwin terminal as I don't have my Linux system right now)

I am getting the following warnings:

common.c:3:1: warning: return type defaults to ‘int’ [-Wreturn-type]
main( )
^
common.c: In function ‘main’:
common.c:6:2: warning: implicit declaration of function ‘workover’ [-Wimplicit-f                  unction-declaration]
workover(i);
^
common.c: At top level:
common.c:9:1: warning: return type defaults to ‘int’ [-Wreturn-type]
workover(i)
^
common.c: In function ‘main’:
common.c:8:1: warning: control reaches end of non-void function [-Wreturn-type]
}
^
  • The first and third say: return type defaults to ‘int’

    which means that if you don't specify the return type, the compiler will implicitly declare it as int

    .
  • The second says, implicit declaration of function ‘workover’

    since the compiler doesn't know what it is workover

    .
  • The third warning is pretty easy to understand and will go away if you fix the first.

You should do it like this:

#include <stdio.h>

int workover(int);

int i;

int main(void)
{
    int i = 5;
    workover(i);
    printf("%d",i);     //prints 5
    return 0;
}

int workover(int i)
{
    i = i*i;    //i will have local scope, so after this execution i will be 25;
    return(i);  //returns 25
}
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