> `uint8_t` and `char` are not guarenteed to be the same type Remember that `char`, `signed char` and `unsigned char` are the different types, even though `cha

> `uint8_t` and `char` are not guarenteed to be the same type Remember that `char`, `signed char` and `unsigned char` are the different types, even though `char` takes the same range of values as either `signed char` or `unsigned char`. The typedef `uint8_t` is usually set to `unsigned char`, not `char`, even on systems where `char` is unsigned. Partly this reflects the fact that `char` is usually used to represent actual characters, while the other two types are usually used to represent integers that take the same amount of memory as `char`. The standard technically does not allow `uint8_t` to be `char` [1], although this requires an extremely pedantic reading. Anyway, if you replace `char` with `unsigned char` in your comment then it's correct. I believe all current major implementations typedef `uint8_t` to `unsigned char`, but that's not guaranteed and even old implementations of GCC had a different type. `unsigned char` satisfies the same relaxed aliasing rule as `char` [2] but `uint8_t` may not. [1] https://stackoverflow.com/a/16002781 https://stackoverflow.com/a/16002781 [2] https://stackoverflow.com/a/40575162 https://stackoverflow.com/a/40575162