On Thu, 29 May 2008 21:05:37 +0000, Jens Thoms Toerring wrote:
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Correct. However, on a machine where CHAR_BIT is anything other than 8,
int8_t cannot be defined. In other words, if int8_t exists, you already
know that CHAR_BIT is equal to 8.
Harald van Dijk <truedfx@gmail. comwrote:
>And when int8_t and int16_t both exist, sizeof(int16_t) cannot possibly
>be anything other than 2,
>be anything other than 2,
But if you have e.g. a machine where CHAR_BITS is 16 wouldn't then
sizeof(int16_t) be 1
sizeof(int16_t) be 1
Correct. However, on a machine where CHAR_BIT is anything other than 8,
int8_t cannot be defined. In other words, if int8_t exists, you already
know that CHAR_BIT is equal to 8.