您如何设置、清除和切换单个位? [英] How do you set, clear, and toggle a single bit?
问题描述
你如何设置、清除和稍微切换?
How do you set, clear, and toggle a bit?
推荐答案
设置一点
使用按位或运算符 (|) 设置位.
number |= 1UL << n;
这将设置 number 的 nth 位.n 应该为零，如果你想设置 1st 位等等直到 n-1，如果你想设置 nth 位.
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 比unsigned long 更宽，则使用1ULL；1UL <<<;n 在评估 1UL << 之后才会发生.n 移动超过 long 的宽度是未定义的行为.这同样适用于所有其他示例.
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.
使用按位与运算符 (&) 清除位.
Use the bitwise AND operator (&) to clear a bit.
number &= ~(1UL << n);
这将清除 number 的 n 位.您必须使用按位非运算符 (~) 反转位字符串，然后将其与.
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 (^) 可用于切换位.
The XOR operator (^) can be used to toggle a bit.
number ^= 1UL << n;
这将切换number的n位.
你没有要求这个，但我不妨加上它.
You didn't ask for this, but I might as well add it.
要检查一点，将数字 n 右移，然后按位与它:
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中.
That will put the value of the nth bit of number into the variable bit.
将 n 位设置为 1 或 0 可以通过以下 2 的补码 C++ 实现来实现:
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);
位n在x为1时置位，x为0时清零.如果 x 有其他值，你会得到垃圾.x = !!x 将其布尔化为 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 的补码否定行为(其中 -1 已设置所有位，与 1 的补码或符号/幅度 C++ 实现不同)，请使用无符号否定.
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 n 位和 (x << n) 会将 nth 位设置为 x.
(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.
通常最好不要复制/粘贴代码，因此很多人使用预处理器宏(例如 社区维基进一步回答)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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