c - 递增一个字节并进行比较，是否定义了此行为？

Question

假设 ab 和 cd 都声明为 unsigned char* 类型。以下代码片段中的行为是否定义明确？我担心的是当 cd[k] = 255 时 1 添加到 cd[k]，预期的行为是比较 ab[k] > cd[k] + 1 将等同于 b[k] > 256？不是 b[k] > 0？这是正确的吗？

if (ab[k] > cd[k] + 1)
{
    r = 1;
}
else if (cd[k] > ab[k] + 1)
{
    r = -1;
}

Answer 1

来自 C11 标准草案 N1570，§6.3.1.1.2:

If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.⁵⁸⁾ All other types are unchanged by the integer promotions.
[...]
⁵⁸⁾ The integer promotions are [...] as part of the usual arithmetic conversions, to certain argument expressions, to the operands of the unary +, -, and ~ operators, and to both operands of the shift operators, as specified by their respective subclauses.

简而言之，每个类型小于 int 或 unsigned int 的表达式都被转换为 int 或 unsigned int，分别。
int 必须至少 16 位宽，因此绝对不会发生回绕，您的表达式将产生 256。