When does uint8_t differ from unsigned char?

When is uint8_t ≠ unsigned char?

According to C and C , CHAR_BIT is guaranteed to be at least 8. This implies that uint8_t, which is defined as a type capable of holding 8-bit values, should be a type with at least 8 bits.

However, a curious situation arises when CHAR_BIT exceeds 8. In such cases, uint8_t can no longer be represented using only 8 bits. This raises the question: can uint8_t be legally defined as a type other than unsigned char on such systems?

C and C

In both C and C , uint8_t must always have the same width as unsigned char. However, it is not required to be the same type. For example, uint8_t could be an extended integer type that is larger than unsigned char and has a different internal representation.

This flexibility in defining uint8_t can provide some advantages. For instance, int8_t is required to be two's complement, while signed char may use ones' complement or sign-magnitude representation. Thus, defining uint8_t as a distinct extended integer type ensures that it consistently follows the two's complement representation.

Additional Considerations

Another potential advantage of defining uint8_t as a distinct type is related to C's aliasing rules. Character types are typically allowed to alias other types, which can hamper compiler optimizations. However, if uint8_t is a distinct non-character type, these aliasing rules will not apply, allowing the compiler to make more aggressive optimizations when working with uint8_t and non-character types concurrently.

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