std :: cout如何在一个补码系统中输出负零? [英] How does std::cout print negative zero in a ones-complement system?

问题描述

在一个补码平台上，以下代码将输出什么?

On a ones-complement platform, what would the following code print?

#include <iostream>

int main() {
    int i = 1, j = -1;

    std::cout << i+j << std::endl;
    return 0;
}

我怀疑它会打印"0"而不是"-0"，但是我似乎找不到任何权威.

I would suspect it would print "0" instead of "-0", but I can't seem to find anything authoritative.

为澄清起见，我对如何打印-0感兴趣，一些人建议在实践中，使用上述代码的情况下，ones-compliment的实现可能不会产生负零.

To clarify, I am interested in how -0 would be printed, several people have suggested that in practice, the implementation of ones-compliment might not generate a negative zero with the above code.

在这种情况下，建议使用以下代码实际生成-0:

In those cases, the following has been suggested to actually generate a -0:

#include <iostream>

int main() {
    std::cout << ~0 << std::endl;
    return 0;
}

问题仍然存在:这将打印什么?

The question still remains: what will this print?

推荐答案

首先，为了说明问题，先使用按位运算来生成负零，然后再使用结果值是不可移植的.就是说，在 fprintf 的文档中(因此，在 std :: basic_ostream :: operator<<(int)的文档中)没有指定符号是否代表int对应于 unsigned 表示形式中的填充位或实际值位.

First of all, just to clarify thing, crafting a negative zero using bitwise operations and then using the resulting value is not portable. That said, nothing specifies in the documentation of fprintf (thus, of std::basic_ostream::operator<<(int)) whether the sign bit in the representation of int corresponds to a padding bit in the representation of unsigned or an actual value bit.

结论是，这是未指定的行为.

As a conclusion, this is unspecified behaviour.

#include <iostream>

int main() {
    std::cout << ~0 << std::endl;
    return 0;
}

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