为什么256是一个signed char在C ++中未定义 [英] Why 256 for a signed char is undefined in C++

问题描述

阅读C ++ Primer第5版的书，我注意到一个 signed char 的值为 256 未定义。
我决定尝试，我看到 std :: cout 没有为那个char变量工作。

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).

但在C上，同样的东西
signed char c = 256;
将为 char c 给出值 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是C ++的情况？

Can someone explain to me why is this the case in C++?

编辑：我知道256是2字节，但是为什么不是C的同样的事情发生在C ++？

I understand that 256 is 2 bytes, but why doesn't the same thing as in C, happen to C++?

推荐答案

编辑：见下面TC的回答。更好。

See T.C.'s answer below. It's better.

签名的整数溢出在C ++和C中未定义。在大多数实现中， signed char ， SCHAR_MAX 是127，因此将256放入其中将溢出它。大多数时候，你会看到数字简单地绕回（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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