为什么我们不能正确地将一个以上的参数传递给extern函数？

Question

I have the following code in my program:

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
int print_numbers_above(double x, float y, int z)
{
    printf("x=%lf, y=%f, z=%d\n", x, y, z);
}
int main()
{
    double x = 1.25;
    float y = 1.25;
    int z = 3;
    print_numbers_below(x, y, z);
    print_numbers_above(x, y, z);
    return 0;
}
int print_numbers_below(double x, float y, int z)
{
    printf("x=%lf, y=%f, z=%d\n", x, y, z);
}

Output:

x=1.25, y=0.000000, z=Garbage

x=1.250000, y=1.250000, z=3

Now, I know that the function print_numbers_below() should have been declared earlier (or should have been defined before main). But it doesn't throw an error. It assumes it to be an extern function (and since it is returning int, there isn't any conflict of return types). Now, I fail to understand why can't I pass more than 1 argument correctly?

(I'm using Visual Studio 2013)

Answer 1

A reasonable compiler would give you the following error:

foo.c:12:5: warning: implicit declaration of function 'print_numbers_below' is invalid in C99 [-Wimplicit-function-declaration]
    print_numbers_below(x, y, z);
    ^
foo.c:16:5: error: conflicting types for 'print_numbers_below'
int print_numbers_below(double x, float y, int z)
    ^
foo.c:12:5: note: previous implicit declaration is here
    print_numbers_below(x, y, z);
    ^

Answer 2

When you call a function the compiler hasn't seen a prototype for, it applies the default argument promotions to all arguments. Chars and shorts are promoted to ints, and floats are promoted to doubles. This breaks when print_numbers_below actually wants a float.

If you haven't defined the function yet, always use a prototype.