当在c中定义和声明的函数原型不同时，为什么没有警告？

Question

I have three files, test.c , ah , ac . test.c call a function declared in ah , and ac define the function. But the funcion in ac are different from ah in the return value and the parameter. Either case, there is no warning from my gcc and there is some result. Why?

In my test.c

#include "a.h"
#include <stdio.h>
int main(){
  int x = a();
  printf("%d\n", x);
}

In my ah

int a();

In my ac

#include <stdio.h>

void a(int a)
{
    printf("%d\n", a);
}

in my terminal:

$ gcc -o test test.c a.c  // no warning
$ ./test
1
2

Answer 1

Without the ah being included in ac, the compiler doesn't know it's a problem. So you won't get any compile issues at all. Each gets its own .o file and everyone is happy.

The compiler expects a() to return an int so it has the machine code grab the result off the stack (or probably some register).

The compiler expects a() to have a parameter, so it has the machine code grab the parameter off the stack (again, probably a register).

Then the linker comes along and puts them all together. It doesn't know that there's an issue, but it makes the call to a() work.

So you get values that are left over on the register (or stack). And hopefully you're in a protected environment so that you're not getting some other user's information.