Is passing additional parameters through function

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Closed 7 years ago.

Possible Duplicate:
Casting a function pointer to another type

Assume i initialize a function pointer with a function that actually takes less parameters then the function pointer definition, will the function still perform correctly if called through the function pointer?

I tried this with gcc and it worked as expected, but i wonder if that behaviour is consistent across compilers/platforms (i suspect in some enviroments it might wreak havoc on the stack):

#include <stdio.h>

typedef void (*myfun)(int, int, int);

void test_a(int x, int y, int z) {
    printf("test_a %d %d %d\n", x, y, z);
}

void test_b(int x, int y) {
    printf("test_b %d %d\n", x, y);
}

int main() {
    myfun fp;
    fp = test_a;
    fp(1, 2, 3);
    fp = (myfun) test_b;
    fp(4, 5, 6);
}

Answer 1

The behavior of your program is undefined. The fact that it compiles at all is because of the cast, which effectively tells the compiler "this is wrong, but do it anyway". If you remove the cast, you'll get the appropriate error message:

a.c:17:8: error: assignment from incompatible pointer type [-Werror]

(From gcc -Wall -Werror.)

More specifically, the behavior depends on the calling conventions. If you were on a platform were the arguments were passed in "reverse" order on the stack, the program would give a very different result.

Answer 2

The function call is undefined behavior.

(C99, 6.3.2.3p8) "[...] If a converted pointer is used to call a function whose type is not compatible with the pointed-to type, the behavior is undefined."

For information note that a function type:

(C99, 6.2.5p20) "[...] describes a function with specified return type. A function type is characterized by its return type and the number and types of its parameters."

Answer 3

It is undefined behavior. Use at your own risk. It has been rumored to cause Nasal Demons!

Answer 4

Whether it works will depend on the calling convention being used.

I wouldn't recommend it.