Fun fact, not: In C, it's difficult to evaluate an expression in 8 bits. Integer promotion implicitly converts almost everything to int. Such as...
void f(uint8_t prev, uint8_t next)
{
if (next == (uint8_t) (prev + 1)) {
/* cast needed to avoid promotion to int */
}
}
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@niconiconi Isn't this technically still an at-least-16-bit operation followed by a cast to uint8 (which is equivalent to an 8bit operation on uint8 in this case)?
@robryk@qoto.org Exactly. But an explicitly cast like this is easily recognized by nearly all compilers to output 8-bit instructions, you don't need a powerful optimizer to do this transformation.