C/C++ packing signed char into int

Question

I have need to pack four signed bytes into 32-bit integral type. this is what I came up to:

int32_t byte(int8_t c) { return (unsigned char)c; }

int pack(char c0, char c1, ...) {
  return byte(c0) | byte(c1) << 8 | ...;
}

is this a good solution? Is it portable (not in communication sense)? is there a ready-made solution, perhaps boost?

issue I am mostly concerned about is bit order when converting of negative bits from char to int. I do not know what the correct behavior should be.

Thanks

Answer 1

char isn't guaranteed to be signed or unsigned (on PowerPC Linux, char defaults to unsigned). Spread the word!

What you want is something like this macro:

#include <stdint.h> /* Needed for uint32_t and uint8_t */

#define PACK(c0, c1, c2, c3) \
    (((uint32_t)(uint8_t)(c0) << 24) | \
    ((uint32_t)(uint8_t)(c1) << 16) | \
    ((uint32_t)(uint8_t)(c2) << 8) | \
    ((uint32_t)(uint8_t)(c3)))

It's ugly mainly because it doesn't play well with C's order of operations. Also, the backslash-returns are there so this macro doesn't have to be one big long line.

Also, the reason we cast to uint8_t before casting to uint32_t is to prevent unwanted sign extension.