std :: Byte位运算符|，＆amp;， ^，〜：为什么要施放无签名的int？
根据std :: byte在 cppReference上的文档a>，用于操作员的实现| std :: byte应该等效于
constexpr std::byte operator|(std::byte l, std::byte r) noexcept
{
return std::byte{ static_cast<unsigned>(l) | static_cast<unsigned>(r) };
}
（操作员＆amp;，^，〜应该类似地实现）
为什么l和r需要将其施放到int int int nunsigned时代码> std :: byte 的基础类型是char unsigned？
注意：我知道char unsigned {} | char unsigned {}在int中结果导致，因为在应用 bitwise或之前将每个操作数升级为int； unsigned {} | unsigned {}返回unsigned，并且没有促销。但是，在这种情况下，我不知道哪些问题可能会导致这样的促销。
According to std::byte's documentation on cppreference, the implementation of operator| for std::byte should be equivalent to
constexpr std::byte operator|(std::byte l, std::byte r) noexcept
{
return std::byte{ static_cast<unsigned>(l) | static_cast<unsigned>(r) };
}
(Operators &, ^, ~ should be implemented similarly)
Why do l and r need to get cast to int unsigned if std::byte's underlying type is char unsigned?
Note: I'm aware that char unsigned{} | char unsigned{} results in an int because each operand gets promoted to int before the bitwise or is applied; while unsigned{} | unsigned{} returns an unsigned and no promotion happens. However, I don't understand which issues may such a promotion cause in this context.
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对于比
int较小的整数类型，在应用操作员之前，它们会晋升为int。这意味着，如果您有一个签名的整数。通过使用
您明确将操作数转换为未签名的整数，以免促销
int发生。在C ++ 20之前，这可能会有所作为，因为与未签名的整数类型不同，不需要使用两者的补体表示形式。
For integer types that are smaller then an
int, they are promoted to anintbefore the operator is applied. That means if you hadthen you would have a signed integer. By using
You explicitly convert the operands to unsigned integers so that no promotion to
inthappens.This could make a difference before C++20 as signed integers were not required to use a two's complement representation unlike unsigned integer types.