c - Does this invoke undefined behaviour?

Question

Consider the following C program:

#include <stdio.h>

int main(){
    int a =-1;
    unsigned b=-1;
    if(a==b)
        printf("%d %d",a,b);
    else
       printf("Unequal");
    return 0;
 }

In the line printf("%d %d",a,b);, "%d" is used to print an unsigned type. Does this invoke undefined behavior and why?

Answer 1

Although you are explicitly allowed to use the va_arg macro from <stdarg.h> to retrieve a parameter that was passed as an unsigned as an int (7.15.1.1/2), in the documentation for fprintf (7.19.6.1/9) which also applies to printf, it explicitly states that if any argument is not the correct type for the format specifier - for an unmodified %d, that is int - then the behaviour is not defined.

As @bdonlan notes in a comment, if the value of b (in this case 2^N - 1 for some N) is not representable in an int then it would be undefined behavior to attempt to access the value as an int using va_arg in any case. This would only work on platforms where the representation of an unsigned used at least one padding bit where the corresponding int representation had a sign bit.

Even in the case where the value of (unsigned)-1 can be represented in an int, I still read this as being technically undefined behavior. As part of the implementation, it would seem to be allowed for an implementation to use built in magic instead of va_args to access the parameters to printf and if you pass something as an unsigned where an int is required then you have technically violated the contract for printf.