如何获得以32位整数为单位的长度

Question

I have tried several variations in JavaScript but none are getting the desired result.

 assert(countIntegerBits(4), 3) // 100 assert(countIntegerBits(8), 4) // 1000 assert(countIntegerBits(20), 5) // 10100 assert(countIntegerBits(100), 7) // 1100100 // https://stackoverflow.com/questions/43122082/efficiently-count-the-number-of-bits-in-an-integer-in-javascript function countIntegerBits(integer) { var length = 0 while (integer = Math.floor(integer)) { if (integer & 1) { length++ } integer /= 2 } return length } function countIntegerBits(integer) { // var length = 0 // while (integer !== 0) { // length += countIntegerBits32(integer | 0) // integer /= 0x100000000 // } // return length // // or perhaps this: // https://gist.github.com/everget/320499f197bc27901b90847bf9159164#counting-bits-in-a-32-bit-integer } function countIntegerBits32(integer) { integer = integer - ((integer >> 1) & 0x55555555) integer = (integer & 0x33333333) + ((integer >> 2) & 0x33333333) return ((integer + (integer >> 4) & 0xF0F0F0F) * 0x1010101) >> 24 } function countStringBits(string) { // looks like this / 8 would be good enough // https://codereview.stackexchange.com/questions/37512/count-byte-length-of-string var length = 0; for (var i = 0; i < normal_val.length; i++) { var c = normal_val.charCodeAt(i); length += c < (1 << 7) ? 1 : c < (1 << 11) ? 2 : c < (1 << 16) ? 3 : c < (1 << 21) ? 4 : c < (1 << 26) ? 5 : c < (1 << 31) ? 6 : Number.NaN } return length; } function countFloatBits(float) { // looks too complicated for an SO question // http://binary-system.base-conversion.ro/real-number-converted-from-decimal-system-to-32bit-single-precision-IEEE754-binary-floating-point.php?decimal_number_base_ten=1.23&sign=0&exponent=01111111&mantissa=00111010111000010100011 } function assert(a, b) { if (a !== b) throw new Error(a + ' != ' + b) } 

What I want to avoid is this convert-to-string hack

var length = integer.toString(2).split('').length

The only other thing I can think of doing is check if bit is set , until you get to the first 1 , then start counting from there.

 assert(countIntegerBits(4), 3) // 100 assert(countIntegerBits(8), 4) // 1000 assert(countIntegerBits(20), 5) // 10100 assert(countIntegerBits(100), 7) // 1100100 function countIntegerBits(integer) { var i = 0 while (true) { if (integer & (1 << i)) { return 31 - i } i++ } } function assert(a, b) { if (a !== b) throw new Error(a + ' != ' + b) } 

But that doesn't seem quite right because I'm not sure if all integers are represented as 32-bits under the hood, for example, (4).toString(2) gives "100" , not 00000000000000000000000000000100 , so not sure.

In there I have explored how to check the length of a string in bits, and same with floats, but while strings seem straightforward if it's utf-8 encoding, it seems like floats is a whole big thing, so my question is only about integers up to the max that JavaScript supports. I will, for all practical purposes for the moment, only be considering integers up to about 1-billion so it doesn't need to account for 123e456 bigints or anything, just numbers up to a few billion, or up to the basic 32-bit integer Max in JavaScript.

Answer 1

There's a relationship between natural log (well, log to any base) and log to another base. For getting the base-2 log:

const log2 = n => Math.log(n) / Math.log(2);

You'd want to round down after adding 1:

const bits = n => Math.floor(log2(n) + 1);