如何在C ++中访问数字的符号位? [英] How can I access the sign bit of a number in C++?

问题描述

我希望能够使用C ++访问数字的符号位.我当前的代码如下所示:

I want to be able to access the sign bit of a number in C++. My current code looks something like this:

int sign bit = number >> 31;

这似乎可行，给我0代表正数，给我-1代表负数.但是，我看不到如何得到负数的-1:如果12是

That appears to work, giving me 0 for positive numbers and -1 for negative numbers. However, I don't see how I get -1 for negative numbers: if 12 is

0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100

那么-12是

1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0011

将其移位31位将使

0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001

它是1，而不是-1，那么为什么在移动它时我得到-1?

which is 1, not -1, so why do I get -1 when I shift it?

推荐答案

在C ++中右移负数的结果是实现定义的.因此，没人知道您的-12右移应该在特定平台上发生什么.您认为它应该具有以上(1)的含义，而我说它可以轻松产生全为模式，即-1.后者称为符号扩展移位.在扩展符号时，符号位被复制到右侧，但从未移出其位置.

The result of right-shifting a negative number in C++ is implementation-defined. So, no one knows what right-shifting your -12 should get on your specific platform. You think it should make the above (1), while I say that it can easily produce all-ones pattern, which is -1. The latter is called sign-extended shifting. In sign-extended shifting the sign bit is copied to the right, but never shifted out of its place.

如果您只对符号位的值感兴趣，那么就不要浪费时间尝试使用按位运算，例如移位等.只需将您的数字与0进行比较，看看它是否为负.

If all you are interested in is the value of the sign bit, then stop wasting time trying to use bitwise operations, like shifts etc. Just compare your number to 0 and see whether it is negative or not.

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