将较小整数的最大值分配给较大整数
考虑以下代码：
uint32_t x = ~uint8_t(0);
std::cout << x << std::endl;
现在，我完全期望它输出 255，但它却输出 4294967295。
我知道 C++ 中的整数提升，但我不明白为什么会发生这种情况。按照我的理解，表达式 ~uint8_t(0) 的计算结果应该是二进制的 1111 1111 。然后，~ 运算符将通过符号扩展值，将类型提升为 int（为了讨论起见，我假设它是 32 位）至0000 0000 0000 0000 0000 0000 1111 1111。然后，应将此提升后的值分配给左值 x，结果为 x == 255。
但显然我没有正确理解这一点。我缺少什么？
Consider the following code:
uint32_t x = ~uint8_t(0);
std::cout << x << std::endl;
Now, I fully expected this to output 255, but instead it output 4294967295.
I'm aware of integer promotion in C++, but I can't understand why this would happen. The way I understand it, the expression ~uint8_t(0) should evaluate to 1111 1111 in binary. The ~ operator will then cause the type to be promoted to an int (which I'll assume is 32-bit for the sake of discussion), by sign extending the value to 0000 0000 0000 0000 0000 0000 1111 1111. This promoted value should then be assigned to the lvalue x, resulting in x == 255.
But obviously I'm not understanding this correctly. What am I missing?
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对一元
~的操作数执行整数提升，因此将uint8_t(0)提升为int，然后将~被评估。它相当于您可以使用 std::numeric_limits::max() 获取类型可表示的最大值;如果类型是无符号的，您还可以将
-1转换为该类型：The integral promotions are performed on the operand of the unary
~, so theuint8_t(0)is promoted tointand then the~is evaluated. It's equivalent toYou can get the maximum value representable by a type using
std::numeric_limits<T>::max(); if the type is unsigned, you can also cast-1to that type:在您的示例中，按位非运算符 (
~) 在执行之前将其操作数提升为int。然后通过赋值将其转换为unsigned int。以下是一些相关的代码示例，可以使这一点更加清晰：
In your example, the bitwise not operator (
~) is promoting its operand to anintbefore it executes. This is then converted to anunsigned intby the assignment.Here are some related code samples that make this more clear: