Type to use to represent a byte in ANSI (C89/90) C?

Question

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

Answer 1

I notice that some answered have re-defined the word byte to mean something other than 8 bits. A byte is 8 bits, however in some c implementations char is 16 bits (2 bytes) or 8 bits (1 byte). The people that are calling a byte 'smallest addressable unit of memory' or some such garbage have lost grasp of the meaning of byte (8 bits). The reason that some implementations of C have 16 bit chars (2 bytes) and some have 8 bit chars (1 byte), and there is no standard type called 'byte', is due to laziness.

So, we should use int_8

Answer 2

Before C99? Platform-dependent code.

But why do you care? Just use stdint.h.

In every implementation of C I have used (from old UNIX to embedded compilers written by hardware engineers to big-vendor compilers) char has always been 8-bit.

Answer 3

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

1. ...
2. The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. [...]
3. 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.

Answer 4

You can find pretty reliable macros and typedefs in boost.

Answer 5

You can always represent a byte (if you mean 8bits) in a unsigned char. It's always at least 8 bits in size, all bits making up the value, so a 8 bit value will always fit into it.

If you want exactly 8 bits, i also think you'll have to use platform dependent ways. POSIX systems seem to be required to support int8_t. That means that on POSIX systems, char (and thus a byte) is always 8 bits.

Answer 6

In ANSI C89/ISO C90 sizeof(char) == 1. However, it is not always the case that 1 byte is 8 bits. If you wish to count the number of bits in 1 byte (and you don't have access to limits.h), I suggest the following:

unsigned int bitnum(void) {
    unsigned char c = ~0u; /* Thank you Jonathan. */
    unsigned int v;

    for(v = 0u; c; ++v)
        c &= c - 1u;
    return(v);
}

Here we use Kernighan's method to count the number of bits set in c. To better understand the code above (or see others like it), I refer you to "Bit Twiddling Hacks".