extern issue

Hi all,

I have worked with extern variable(s) a little. But, how does the extern function(s) work. Here is a sample code which i recently saw. Can anyone tell me why the compiler is not complaining about the usage and how does this code work

Test1.c file source code

Code:

//test1.c
#include <stdio.h>

void dummy(int i,int j)
{
        printf("I am inside the dummy function with value %d and %d\n",i,j);
}

Test2.c file source code

Code:

//test2.c
int main()
{
        extern void dummy(int i); //Here the arguments are mismatched

        dummy(2);
        return 0;
}

Compiled it using the following gcc command

gcc -Wall -o extern test1.c test2.c

and received not even a warning and when i tried to run this code

./extern
I am inside the dummy function with value 2 and -1075886112

How is this working. Any clues.

I think the result is undefined. That is; it may very well crash your application...

Thanks EVOEx, but some of my friends suggest that extern functions act like function pointers. Any idea on that?

"extern" on a function prototype is useless, since function prototypes are extern by default (unless they're inline). Notice that in your example, if you remove "extern" from your example it compiles exactly the same and works exactly the same. C can't save you if you lie to it with the prototypes.

tabstop thanks for your explanation.