Question

I ran across the following example program and I don’t exactly understand it’s output:

#include <stdio.h>

int main( void ) {

    unsigned char i, m =0xFF, n=0x1;

    for ( i = 0; i != 8; i++,n+=n, m/=2 )
        printf("%5x %5x %5x %5x %5x %5x\n", n,m,n&m,n|m,n^m,~n);

    return 0;
}

It prints out:

    1    ff     1    ff    fe fffffffe
    2    7f     2    7f    7d fffffffd
    4    3f     4    3f    3b fffffffb
    8    1f     8    1f    17 fffffff7
   10     f     0    1f    1f ffffffef
   20     7     0    27    27 ffffffdf
   40     3     0    43    43 ffffffbf
   80     1     0    81    81 ffffff7f

The problem is that last column. Since it’s unsigned char I would expect it to print out just 2 places in every column. ~n produces an unsigned char as it’s result, but it seems like it’s being cast to a signed 32 bit value and sign extended by the %5x specifier.

How is that possible, what’s going on here?

Answer 1

Integer types are promoted when they are used in arithmetic operations (this has nothing to do with printf, by the way).

So, for example,

unsigned char x = 0xff;
int y = ~x; // x is promoted to 0x000000ff, then changed to 0xffffff00
unsigned char z = ~x; // truncated back to 0x00

Integer promotion causes various problems:

unsigned char x = 1;
if (x << 8)
    puts("x << 8 is true"); // does print
x <<= 8;
if (x)
    puts("x <<= 8 is true"); // does not print

The two ways to truncate things are casting and masks. Use whatever you prefer.

unsigned char x = 0xab;
printf("x = %02x\n", (unsigned char) x);
printf("x = %02x\n", x & 0xff);

Integer promotion doesn’t always happen, and it’s not the only kind of implicit cast. It’s also a bit subtle and the exact rules are difficult to remember. You only really need to worry about it if you’re working with 64-bit numbers, because 1U << 32 could end up being 0 or 1 or something else entirely. (It’s often 1 on x86).