如何设置警戒/解除警戒位在漫长的立场呢？ [英] How to set/unset a bit at specific position of a long?

问题描述

如何设置警戒/解除警戒位为长期在Java中的特定位置？

How to set/unset a bit at specific position of a long in Java ?

例如，

long l = 0b001100L ; // bit representation

我要在位置2和取消位3位设置位从而相应的长会，

I want to set bit at position 2 and unset bit at position 3 thus corresponding long will be,

long l = 0b001010L ; // bit representation

任何人可以帮助我该怎么做？

Can anybody help me how to do that ?

推荐答案

要设置位，用途：

x |= 0b1; // set LSB bit
x |= 0b10; // set 2nd bit from LSB

要擦除位使用：

x &= ~0b1; // unset LSB bit (if set)
x &= ~0b10; // unset 2nd bit from LSB

切换了一下使用方法：

to toggle a bit use:

x ^= 0b1;

请注意我用0B？你也可以使用任意整数，例如：

Notice I use 0b?. You can also use any integer, eg:

x |= 4; // sets 3rd bit
x |= 0x4; // sets 3rd bit
x |= 0x10; // sets 9th bit

然而，这使得它更难知道哪些位被改变。

However, it makes it harder to know which bit is being changed.

使用二进制可以让你看到它的确切位将被设置/删除/切换。

Using binary allows you to see which exact bits will be set/erased/toggled.

要动态地设置位，用途：

To dynamically set at bit, use:

x |= (1 << y); // set the yth bit from the LSB

（1 LT;＆LT; Y）。的移动左... 001 y位，这样你就可以移动设置位y地方

(1 << y) shifts the ...001 y places left, so you can move the set bit y places.

您也可以一次设定多个位：

You can also set multiple bits at once:

x |= (1 << y) | (1 << z); // set the yth and zth bit from the LSB

还是要取消设置：

Or to unset:

x &= ~((1 << y) | (1 << z)); // unset yth and zth bit

或切换：

x ^= (1 << y) | (1 << z); // toggle yth and zth bit

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