按位运营商简单地翻转所有位整数？ [英] Bitwise operator for simply flipping all bits in an integer?

问题描述

我要翻转所有位整数的二进制重新presentation。鉴于：

I have to flip all bits in a binary representation of an integer. Given:

10101

输出应

01010

什么是位运算符与一个整数使用时，做到这一点？例如，如果我在写像 INT flipBits（INT N）的方法; ，你会在身上去了？我需要翻转只有什么是已经present的数量，而不是在所有的整数32位。

What is the bitwise operator to accomplish this when used with an integer? For example, if I were writing a method like int flipBits(int n);, what would go in the body? I need to flip only what's already present in the number, not all 32 bits in the integer.

推荐答案

的〜单目运算符是按位否定。如果你需要比在什么适合较少的比特 INT 那么你就需要使用＆放大器来掩盖它;之后事实。

The ~ unary operator is bitwise negation. If you need fewer bits than what fits in an int then you'll need to mask it with & after the fact.

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