I have a "boolean bit array",
const array: boolean[] = [false, true, false, true]; // 0101
How can I get a number 5? Thanks
5 Answers
I don't know TS, in pure JS it's
a = [false, true, false, true]
b = a.reduce((res, x) => res << 1 | x)
alert(b)To do the opposite (i.e. number to array):
b = 5
a = b ? [] : [false]
while(b) {
a.push((b & 1) === 1)
b >>= 1
}
alert(a)or
b = 5
a = b.toString(2).split('').map(x => x === '1');
alert(a)5 Comments
Hongbo Miao
res << 1 | x, 0, what is this , 0? Because if I change to res << 1 | x, it still works. thanksHongbo Miao
thanks again, how can I reserve the first
false too? right now the result is true,false,true. How can I control the size of output array easily, like if I give a 4, it shows false,true,false,true; if I give a 5, it shows false,false,true,false,true
georg
@HongboMiao: just add as many falses as you want:
while a.length < desired a.unshift(false).
xinthose
I really do not understand what is happening in the reduce function
(res, x) => res << 1 | x. I know it is a bitwise OR operation.georg
@xinthose: it shifts the accumulator to the left by one and sets the rightmost bit to
x (1 or 0). For example, let res=6 (110) and x = 1, then res<<1 is 12 (1100) and res<<1|x is 13 (1101)This works for me with typescript.
async maskBoolToInt(boolArray:boolean[]){
let debugmode = true;
if(debugmode){
console.log('Debug : "maskBoolToInt" Started');
console.log('boolArray = ' + boolArray);
}
let bitArray:number[] = [];
boolArray.forEach((element) => {
bitArray.push(+element); //convert bool to bit
});
if(debugmode){
console.log('bitArray = ' + bitArray);
}
let result: any = bitArray.reduce((accumulator: number, currentValue: number) => accumulator << 1 | currentValue); //bitwise conversion to integer
if(debugmode){
console.log('result = ' + result);
console.log('Debug : "maskBoolToInt" Finished');
}
return result
};
Comments
I would use simple number/radix with string split/join function to do that.
const numberToBoolArr = (n: number): Array<boolean> => (n).toString(2).split('').map(r => r === '1')
const boolArrToNumber = (arr: Array<boolean>): number =>
parseInt(arr.map(r => r ? '1' : '0').join(''), 2)
With the boolArrToNumber, you can verify that:
console.log(boolArrToNumber([false, true, false, true])) // 5
Comments
Doesn't answer the question, but touches on the related topic of representing an array of boolean values as a number and converting it back.
const boolsToNum = (bools: boolean[]) => {
return bools.reduceRight((res, bool) => res << 1 | +bool, 1)
}
const numToBools = (num: number) => {
const bools = []
while (num > 1) {
bools.push((num & 1) === 1)
num >>= 1
}
return bools
}
reduceRight() is used instead of reduce() to remove the need of reversing the array when converting a number back to bools. Its initial value used as 1 instead of 0 to preserve the size of array and to keep array starting false values. It works as if we passed in the array of +1 length, which first value is true. Costs 1 bit but removes the need of checking array length later. This bit is dropped when converting back by while (num > 1)
const array:Array<boolean> = [false, true, false, true]; // 0101
console.log(array, 'original')
const num = boolsToNum(array)
console.log(num, 'compressed')
console.log(numToBools(num), 'uncompressed')
// (4) [false, true, false, true] original
// 26 compressed
// (4) [false, true, false, true] uncompressed
Comments
Add a naive solution
const booleanArrayToInt = (array: boolean[]) => {
let res = 0;
for (let i = 0; i < array.length; i++) {
if (array[array.length - i - 1]) {
res += 2 ** i;
}
}
return res;
};
console.log(booleanArrayToInt([false, true, false, true])); // 5
parseInt(array.map(i => i+0).join(''), 2)//5