c++ - 负尺寸_t

Question

是否明确规定(对于一般的无符号类型)，即:

static_assert(-std::size_t{1} == ~std::size_t{0}, "!");

我刚刚研究了 libstdc++ 的 std::align 实现并注意使用 std::size_t 取反:

inline void*
align(size_t __align, size_t __size, void*& __ptr, size_t& __space) noexcept
{
  const auto __intptr = reinterpret_cast<uintptr_t>(__ptr);
  const auto __aligned = (__intptr - 1u + __align) & -__align;
  const auto __diff = __aligned - __intptr;
  if ((__size + __diff) > __space)
    return nullptr;
  else
    {
      __space -= __diff;
      return __ptr = reinterpret_cast<void*>(__aligned);
    }
}

Answer 1

无符号整数类型被定义为环绕，无符号整数类型中可表示的最大可能值是所有位都设置为 1 的数字 - 所以是的。

如 cpp-reference 所述(arithmetic operators / overflow):

Unsigned integer arithmetic is always performed modulo 2ⁿ where n is the number of bits in that particular integer. E.g. for unsigned int, adding one to UINT_MAX gives ​0​, and subtracting one from 0​ gives UINT_MAX.

相关:Is it safe to use negative integers with size_t?