乘以 unsigned Short 会导致未定义的行为吗？

作为“的后续行动”https://stackoverflow.com/questions/33732041/why-static-castunsigned-intushrt-maxushrt-max-yields- Correct-value https://stackoverflow.com/questions/33732041/why-static-castunsigned-intushrt-maxushrt-max-yields-correct-value"

我问自己是否要推广排名低于以下的所有类型（除了一些例外）int to int执行算术运算在某些情况下可能会导致 UB。

e.g.:

unsigned short a = 0xFFFF;
unsigned short b = a*a;

由于无符号短被提升为int对于算术运算，这将导致：

unsigned short a = 0xFFFF;
unsigned short b = (int)a*(int)a;

As (int)0xFFFF*(int)0xFFFF导致溢出，有符号类型的溢出为UB：可以将两个无符号的 Short 相乘x,y在这种情况下导致未定义的行为x*y > INT_MAX

UPDATE:

该问题具体针对以下情况int是 32 位并且short是16位的。

” Unsigned integers, declared unsigned, shall obey the laws of arithmetic modulo 2ⁿ where n is the number of bits in the value representation of that particular size of integer.

Apart from the slightly misleading wording about “declared unsigned” this might seem to apply that every arithmetic expression that involve only argument of some given unsigned type, will yield a result modulo 2ⁿ for that type.

However, there are no arithmetic expressions at all for unsigned types of lower conversion rank than int: all arguments in an apparent such expression are converted up to ⁽¹⁾at least int, or depending on the number ranges of the C++ implementation, up to unsigned int.

As a result, a*b where a and b are unsigned short values, ⁽²⁾can have formally Undefined Behavior. Because it's not an unsigned short expression. It's (in practice) an int expression.

也就是说，使用一个合理的编译器，不会在注意到正式 UB 的地方引入特殊大小写，并且在实践中使用 8 位字节和unsigned short可以表示的最大值int，以及常见的二进制补码有符号整数表示，结果转换回unsigned short， 将as if它是范围内的模算术unsigned short。这是因为在机器代码级别，二进制补码只是以 0 为中心范围的模算术。

^{(1) In practice one will usually be using an 8 bits-per-byte implementation where the maximum value of unsigned short fits well within the int range, so in practice we're talking about a conversion up to int.
(2) E.g., for 16-bit unsigned short and 32-bit int, (216−1)2 = 232−2×216+1 > 231−1, where the last value is the maximum positive int value.}