c - unsigned char 的隐式转换(算术转换与赋值)

Question

在做某事时遇到了类似于以下的代码:

#define MODULUS(a,b)        ((a) >= 0 ? (a)%(b) : (b)-(-a)%(b))

unsigned char w;
unsigned char x;
unsigned char y;
char z;

/* Code that assigns values to w,x and y.  All values assigned 
   could have been represented by a signed char. */

z = MODULUS((x - y), w);

据我了解，算术 (x - y) 将在任何类型转换之前完成，并且宏将始终计算为 (a)%(b) - 结果将是一个始终大于或等于零的无符号字符。然而，代码按预期运行，我认为我的理解是有缺陷的。所以...

我的问题是:

1. 在计算表达式之前是否会进行到有符号字符的隐式类型转换？

2. 是否存在上述代码不起作用的情况(例如，如果无符号值足够大以至于无法用有符号值表示)？

Answer 1

Does an implicit type conversion to signed char occur before the expression is evaluated?

不，到 int 的转换发生在表达式 x - y 计算之前。因此结果可能为负。

Is there a situation (for example, if the unsigned values were large enough that they could not be represented by a signed value) where the above code would not work?

如果sizeof int == 1，整数提升会将unsigned char提升为unsigned int，这可能会产生错误结果是因为在按 w 取模之前，由于无符号算术而执行了按 UINT_MAX + 1 取模。

¹ 默认整数提升。