3 ms·
I thought if uint8_t exists, it logically must be the same as unsigned char? You put some of the statements of the standards together and that's their combined
by kzrdude 3y ago
I thought if uint8_t exists, it logically must be the same as unsigned char? You put some of the statements of the standards together and that's their combined consequence.
- quietbritishjim 3y agoNo it doesn't have to be, and it wasn't. That was the whole problem. They were both unsigned integer types with 8 bit widths but they were distinct and, in particular, uint8_t didn't have the aliasing exception that unsigned char does.