Is it better to use uint8_t or char to represent binary data arranged in bytes?

Question

I like uint8_t because I feel that the intent (binary data, not characters) is better expressed. However, there are many char-based I/O functions which I might want to call. Just wondering if there's a best practice that folks conform to out there? This is my first foray into byte-level I/O.

Answer 1

The problem with char is that it may be either a signed type (with a range of at least -127..+127) or an unsigned type (with a range of at least 0..255).

unsigned char is better than plain char for binary byte-oriented data.

uint8_t, if it exists, will have the same range and representation as unsigned char, and will probably be just another name for the same type.

Note that C++ (like C) defined a byte as the smallest addressible storage unit on a given system. It's at least 8 bits, and will probably be exactly 8 bits on any system you're likely to use, but it's allowed to be wider. The macro CHAR_BIT, defined in <stddef.h> or <cstddef>, tells you the number of bits in a byte.

If you want to absolutely guarantee that you're using 8-bit bytes, either query the value of CHAR_BIT or use uint8_t. If bytes are bigger than 8 bits, then uint8_t won't exist (because there will be no type that meets its requirements).