Thread: Unexpected behavior of stdarg

Here's a small test program that demonstrates what I'm experiencing. I'm building it with gcc 4.2.1.

Code:

#include <stdarg.h>                                                 
#include <stdlib.h>                                                 
#include <stdio.h>                                                  
                                                                    
void test(int foo, ...) {                                           
        va_list args;                                               
        va_start(args, foo);                                        
        printf("%lld\n", va_arg(args, long long));                  
        va_end(args);                                               
}                                                                   
                                                                    
int main() {                                                        
        // I would expect this to output -1                         
        test(0, -1);                                                
                  
        // Works with a cast                                         
        test(0, (long long)-1);                                     

        return 0;                                                   
}

The output of this is:

4294967295
-1

The first number corresponds to (unsigned)-1. I cannot figure out why this conversion is being made. My understanding (and probably incorrect) is that the compiler should promote the untyped value of -1 to the type specified in va_arg.

Can anyone shed light on why this is happening?

Thanks,

Fred

Question 12.9
That's why it's *WRONG* to pass 0 as NULL pointer to vararg function
Even though you can pass 0 as NULL pointer to function with fixed args without any problem.
IMO, you can simply just try printf("%f\n", 3);