How to find Bit Difference between two numbers in Javascript

Question

假设我有2个数字，例如1和2。它们的二进制表示形式是'01'和'10'，所以它们的位差是2。对于数字5和7，二进制表示形式是'101'和'111'，所以位差是1当然，我可以将两个数字都转换为二进制，然后循环查找其差值，但是有没有更简单的方法呢？

Answer 1

Ah, a string op is an easy way to do this per CertainPerformance's answer, but here's a purely bitwise solution.

This is a flattened loop that handles JavaScript's limited 32-bit int support.

 // implementation of the "bit population count" operation for 32-bit ints function popcount(u) { // while I'm at it, why not break old IE support :) if ( !Number.isInteger(u) ) throw new Error('Does not actually work with non-integer types.'); // remove the above check for old IE support u = (u & 0x55555555) + ((u >> 1) & 0x55555555); u = (u & 0x33333333) + ((u >> 2) & 0x33333333); u = (u & 0x0f0f0f0f) + ((u >> 4) & 0x0f0f0f0f); u = (u & 0x00ff00ff) + ((u >> 8) & 0x00ff00ff); u = (u & 0x0000ffff) + ((u >>16) & 0x0000ffff); return u; } // select all bits different, count bits function diffcount(a, b) { return popcount( a ^ b ); } // powers of two are single bits; 128 is common, bits for 16, 32, and 8 are counted. // 128+16 = 144, 128+32+8 = 168 console.log(diffcount(144,168)); // = 3 // -1 is 4294967295 (all bits set) unsigned console.log(diffcount(-1,1)); // = 31 // arbitrary example console.log(diffcount(27285120,31231992)); // = 14 

If you need arbitrarily large values, let me know...

It'll require the use of typed arrays, the above functions, and a loop.

Answer 2

You could use bitwise XOR ( ^ ) to identify the positions whose bits are different, convert the result to a string, then count up the number of occurrences of 1 in the string:

 const bitDiffCount = (a, b) => { const bitStr = ((a ^ b) >>> 0).toString(2); return bitStr.split('1').length - 1; }; console.log(bitDiffCount(5, 7)); console.log(bitDiffCount(1, 2)); console.log(bitDiffCount(16, 16)); console.log(bitDiffCount(16, 17)); console.log(bitDiffCount(16, 18)); console.log(bitDiffCount(16, 19)); 

Answer 3

1) XOR both the numbers: x = a^b .

2) check if the XOR result is a power of 2. if it's power of 2 then there's only 1-bit difference. (x && (!(x & (x - 1)))) should be 1

bool differAtOneBitPos(unsigned int a, 
                       unsigned int b) 
{ 
    return isPowerOfTwo(a ^ b); 
}

bool isPowerOfTwo(unsigned int x) 
{ 
    return x && (!(x & (x - 1))); 
}