[DSMan195276]

> Also, apparently uint8_t may not be a character type.

I think that goes hand-in-hand with the fact that `char` is not guaranteed to be 8-bits wide in C, so by that fact `uint8_t` may not be a `char`. In practice I highly doubt this distinction really matters: Platforms without an 8-bit `char` are basically guaranteed to not support (a standards compliant) `uint8_t` anyway, and it is reasonable to target 8-bit `char` systems in which case it's safe to assume `uint8_t` and `char` are the same thing. (Well, `unsigned char`)

[TorKlingberg]

Yes, a system with char >8 bits wouldn't have a uint8_t type at all. But, even when char and (u)int8_t are the same size, it may not be the same type. Compilers could go and apply strict aliasing for pointers to uint8_t. I'm at the limit of my C types knowledge here, so correct me if I'm just wrong.