Type to use to represent a byte in ANSI (C89/90) C?
c, c89, types
Solution
`char` is always a byte , but it's not always an octet. A byte is the smallest addressable unit of memory (in most definitions), an octet is 8-bit unit of memory.
That is, `sizeof(char)` is always 1 for all implementations, but `CHAR_BIT` macro in `limits.h` defines the size of a byte for a platform and it is not always 8 bit. There are platforms with 16-bit and 32-bit bytes, hence `char` will take up more bits, but it is still a byte. Since required range for `char` is at least -127 to 127 (or 0 to 255), it will be at least 8 bit on all platforms.
ISO/IEC 9899:TC3
6.5.3.4 The sizeof operator
- ...
- The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. [...]
- When applied to an operand that has type `char`, `unsigned char`, or `signed char`, (or a qualified version thereof) the result is 1. [...]
Emphasis mine.
Problem
Is there a standards-complaint method to represent a byte in ANSI (C89/90) C? I know that, most often, a char happens to be a byte, but my understanding is that this is not guaranteed to be the case. Also, there is stdint.h in the C99 standard, but what was used before C99? I'm curious about both 8 bits specifically, and a "byte" (sizeof(x) == 1).