为什么在 C++ 中未定义有符号字符的 256
阅读 C++ Primer 5th edition book，我注意到一个值为 256 的 signed char 是未定义的.我决定尝试一下，我发现 std::cout 不适用于该 char 变量.(无印刷).
Reading the C++ Primer 5th edition book, I noticed that a signed char with a value of 256 is undefined.
I decided to try that, and I saw that std::cout didn't work for that char variable. (Printed Nothing).
但是在 C 上，同样的事情signed char c = 256;将为 char c 提供一个值 0.
But on C, the same thing
signed char c = 256;
would give a value 0 for the char c.
我尝试搜索但没有找到任何东西.
I tried searching but didn't find anything.
有人可以向我解释为什么在 C++ 中会出现这种情况吗?
Can someone explain to me why is this the case in C++?
我知道 256 是 2 个字节，但为什么 C++ 中的情况与 C 不同?
I understand that 256 is 2 bytes, but why doesn't the same thing as in C, happen to C++?
推荐答案
见下面 T.C. 的回答.这样更好.
See T.C.'s answer below. It's better.
有符号整数溢出在 C++ 和 C 中是未定义的.在大多数实现中，signed char 的最大值 SCHAR_MAX 为 127，因此将 256 放入其中会溢出它.大多数情况下，您会看到数字只是环绕(到 0)，但这仍然是未定义的行为.
Signed integer overflow is undefined in C++ and C. In most implementations, the maximum value of signed char, SCHAR_MAX, is 127 and so putting 256 into it will overflow it. Most of the time you will see the number simply wrap around (to 0), but this is still undefined behavior.
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