如何设置，清除和切换单个位? [英] How do you set, clear, and toggle a single bit?

问题描述

如何设置，清除和切换位?

How do you set, clear, and toggle a bit?

推荐答案

设置位

使用按位或运算符(|)设置一个位.

number |= 1UL << n;

这将设置number的第n位.如果要设置1 st位，将n设置为零；如果要设置n th位，将n-1设置为n-1.

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

如果number宽于unsigned long，请使用1ULL；否则，请使用1ULL.直到评估了1UL << n之后，1UL << n的提升才发生，在这种情况下，未定义的行为会移动超过long的宽度.其余所有示例都一样.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

使用按位AND运算符(&)清除一位.

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

这将清除number的第n位.您必须使用按位NOT运算符(~)反转位字符串，然后将其取反.

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

XOR运算符(^)可用于切换一位.

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

这将切换number的第n位.

您没有要求这样做，但我也可以添加它.

You didn't ask for this, but I might as well add it.

要检查一下，将数字n右移，然后按位与它相符:

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

这会将number的第n位的值放入变量bit.

That will put the value of the nth bit of number into the variable bit.

使用2的补码C ++实现，可以将第n位设置为1或0:

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

如果x是1，则

位n将被置位，如果x是0，则清零.如果x具有其他值，则会产生垃圾. x = !!x会将其布尔值设置为0或1.

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

要使其独立于2的补码取反行为(其中-1设置了所有位，与1的补码或符号/幅值C ++实现不同)，请使用无符号取反.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

或

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

使用无符号类型进行可移植位操作通常是一个好主意.

It's generally a good idea to use unsigned types for portable bit manipulation.

或

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n))将清除第n位，而(x << n)将第n位设置为x.

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

通常不要复制/粘贴代码也是一个好主意，因此很多人使用预处理器宏(例如社区Wiki进一步回答)或某种封装.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.

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