The behavior of a C compiler with old-styled funct

Question

When my program consists of two files:

main.c

#include <stdio.h>

int main(void) { 
     printf("%lf\n",f());   
     return 0;
 }

func.c

double f(int a) {
 return 1;
}

compiler do not show any errors.

When my program consists of only one file:

main.c

#include <stdio.h>

int main(void) { 
     printf("%lf\n",f());   
     return 0;
 }

double f(int a) {
 return 1;
}

Visual C++ 2008 compiler show the following error:

Error 2 error C2371: 'f' : redefinition; different basic types d:\temp\projects\function1\function1\1.c 8 function1

Can anybody explain this strange behavior?

Answer 1

Your first example never uses func.c so I'm not sure exactly what the compiler is doing about f() because it has no definition.

In the second example, I don't know why you can't have two functions with different signatures, but you aren't calling the function you've defined. You call f() with no arguments, but the f you define takes an int which makes it a different function.

Answer 2

Both the programs are wrong.

Without a prototype in scope, a compiler assumes that a function returns int and takes an unspecified number of parameters.

Let's change your files a bit:

$ cat func.c
double f(int a) {
    return 1.0;
}
$ cat main.c
#include <stdio.h>

int main(void) { 
    double d = f();
    printf("%lf\n", d);
    return 0;
}

When I compile it, gcc warns me (Visual C++ should too, in conformant mode). But let's ignore the warning.

$ gcc -std=c99 -pedantic -W -Wall func.c main.c -o test
func.c:1: warning: unused parameter 'a'
main.c: In function 'main':
main.c:4: warning: implicit declaration of function 'f'
$ ./test
0.000000

It did not print 1, but printed 0. This is because the compiler assumed that f() returned an int, and the assignment d = f(); converted that "int" to a double. The compiler still compiled the code because it couldn't tell that f() wasn't defined the way it was (implicitly) declared. But compiling the above program isn't required by the standard, so the compiler could have rejected it (try with gcc -Werror for example!)

If we have everything in one file:

$ cat func.c >>main.c
$ gcc -std=c99 -pedantic -W -Wall func.c main.c -o test
main.c:4: warning: implicit declaration of function 'f'
main.c: At top level:
main.c:9: error: conflicting types for 'f'
main.c:4: error: previous implicit declaration of 'f' was here
main.c:9: warning: unused parameter 'a'

Now the compiler sees the conflict, and gives you an error message. But, a compiler is not required to reject the above program, it may or may not.

Most compilers don't reject the first program because they don't know if you have a correct definition of the function f() in another translation unit or not. They reject the second program because they know that you don't.

Answer 3

C will assume a function has the prototype int func(); unless you have told it otherwise.(Note that in C int func(); and int func(void); are different things)

In your second case, you do a call to f() for which the compiler hasn't seen any prototype, so it assumes it is int f(); . Later on it sees your definition for f() which has a different prototype - and issues an error.

This doesn't happen in the 1. case, as they're in different compilation units.