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问题描述
如果我有一个基本的位掩码......
If I have a basic bitmask...
cat = 0x1;
dog = 0x2;
chicken = 0x4;
cow = 0x8;
// OMD has a chicken and a cow
onTheFarm = 0x12;
...如何检查是否只设置了一只动物(即一位)?
...how can I check if only one animal (i.e. one bit) is set?
onTheFarm
的值必须是2 ,但如何以编程方式检查(最好用Javascript)?
The value of onTheFarm
must be 2, but how can I check that programmatically (preferably in Javascript)?
推荐答案
您可以使用此代码计算在非负整数值中设置的位数(改编自):
You can count the number of bits that are set in a non-negative integer value with this code (adapted to JavaScript from this answer):
function countSetBits(i)
{
i = i - ((i >> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
return (((i + (i >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24;
}
它应该比单独检查每个位更有效。但是,如果在 i
中设置了符号位,则它不起作用。
It should be much more efficient than examining each bit individually. However, it doesn't work if the sign bit is set in i
.
编辑(所有信用到Pointy的评论):
EDIT (all credit to Pointy's comment):
function isPowerOfTwo(i) {
return i > 0 && (i & (i-1)) === 0;
}
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