/*
    printing-bytes.c
    Jeff Ondich, 23 Jan 2022

    An exploration of the arduous business of printing a single byte
    as a two-digit hexadecimal number. C's automatic typecasting causes
    surprising problems.
 */

#include <stdio.h>

int main(int argc, char *argv[]) {
    // Various ways of printing a char's value in hexadecimal. At no time
    // do the bits in the relevant byte change, but several of these
    // printing options give us an answer we don't want.
    //
    // printf expects an int for %x. If it sees something that isn't
    // already an int (e.g. a char), printf does typecasting to transform the
    // argument to an int. And when a char gets typecast to an int, the
    // additional 24 bits get filled with whatever the leftmost bit of the
    // char is. So if the leftmost bit of the char is 1, then the int ends up as
    // a negative number, and in hex that looks like 0xFFFFFF?? where
    // the ?? is the contents of the original char.
    char ch = 0xc3;
    int n = (int)ch;
    printf("0x%x\n", n);
    printf("0x%x\n", ch);
    printf("0x%x\n", (unsigned char)ch); // yay! this one does what we want!
    printf("0x%x\n", (unsigned int)ch);
    printf("0x%x\n", ch & 0xFF); // yay! this one too!

    // Stay tuned. We're going to look at the assembly generated by this
    // program to figure out why the heck it works this way.

    return 0;
}