3 ms·

Probably not a big deal assuming uint8_t is a typedef for unsigned char.

• beefhash 8y ago
  The problem is that this is an assumption, not a guarantee. If somewhere, somewhen, some toolchain decides "actually, let's define this to __u8 and do fancy stuff with this compiler-internal type", your code breaks in the most mysterious way possible.
  • v_lisivka 8y ago
    Type uint8_t is guaranteed to have 8 bit, no padding bits, 2s complement. You are talking about uint_least8_t or int_fast8_t. Quote: Exact-width integer types The typedef name intN_t designates a signed integer type with width N, no padding bits, and a two's-complement representation. Thus, int8_t denotes a signed integer type with a width of exactly 8 bits. The typedef name uintN_t designates an unsigned integer type with width N. Thus, uint24_t denotes an unsigned integer type with a width of exactly 24 bits.
    • Mindless2112 8y ago
      But what about unsigned char? It is guaranteed to have CHAR_BIT bits, where CHAR_BIT is at least 8. If CHAR_BIT is greater than 8, then uint8_t cannot be typedef'd to unsigned char. I've stopped programming in C if I can help it. It's a crazy language with support for crazy machines.
      • TheCoelacanth 8y ago
        If CHAR_BITS is greater than 8, then uint8_t cannot be defined because CHAR_BITS is defined as the width of the smallest object that is not a bit-field. If uint8_t exists, it must be the same width as unsigned char.
        • beefhash 8y ago
          It must have the same width as unsigned char in that scenario, but that does not make it unsigned char. § 6.5 of C11 only allows casting to "a character type" (and a few cases irrelevant in this context), not "a type of CHAR_BITS width". There's no requirement that uint8_t is a typedef for unsigned char.