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This is an odd one, but stick with me here. Given a hex value (from 00-FF), I need to convert it into binary, then from there convert it into an object with the key being the 'place' of each bit, and the value being a boolean.

That was an awkward explanation. Here's an example:

Given 'FF', I need to translate that into binary (parseInt(<hex value>, 16).toString(2) works for that job).

Now that FF has bee 1111 1111, I need to convert that into an object that looks like so:

{ 
  "1": true, 
  "2": true, 
  "4": true, 
  "8": true, 
  "16":true, 
  "32":true,
  "64": true, 
  "128": true
}

As another example, given the hex value 'A6', that bees 1010 0110 which would bee:

{ 
  "1": false, 
  "2": true, 
  "4": true, 
  "8": false, 
  "16":false, 
  "32":true, 
  "64": false, 
  "128": true
}

The only thing I'm looking to do is that final conversion from binary -> object.

I know I can do it with a simple loop, but I was wondering if there's any cool p-sci way to acplish this.

Thanks!

This is an odd one, but stick with me here. Given a hex value (from 00-FF), I need to convert it into binary, then from there convert it into an object with the key being the 'place' of each bit, and the value being a boolean.

That was an awkward explanation. Here's an example:

Given 'FF', I need to translate that into binary (parseInt(<hex value>, 16).toString(2) works for that job).

Now that FF has bee 1111 1111, I need to convert that into an object that looks like so:

{ 
  "1": true, 
  "2": true, 
  "4": true, 
  "8": true, 
  "16":true, 
  "32":true,
  "64": true, 
  "128": true
}

As another example, given the hex value 'A6', that bees 1010 0110 which would bee:

{ 
  "1": false, 
  "2": true, 
  "4": true, 
  "8": false, 
  "16":false, 
  "32":true, 
  "64": false, 
  "128": true
}

The only thing I'm looking to do is that final conversion from binary -> object.

I know I can do it with a simple loop, but I was wondering if there's any cool p-sci way to acplish this.

Thanks!

• javascript

Share Improve this question Follow asked May 31, 2016 at 22:11 Kieran EKieran E 3,67633 gold badges2121 silver badges4343 bronze badges 3

• 1 I can think of many cool more plicated ways to do it, but a loop seems in order in this case. Just don't calculate the numeric keys every time, do it once and store them. – Eran Goldin Commented May 31, 2016 at 22:16
• 2 Note that you don't need to convert the value into a string: x = parseInt('A6', 16) // 255; for (n = 1; x > 0; x = x >> 1, n = n * 2) { console.log(n, x % 2 == 1) } – Hamms Commented May 31, 2016 at 22:23
• @Hamms You should really post that (with a little extra) as an answer. It's trivial to go from that to the object OP wants to generate. – Mike Cluck Commented May 31, 2016 at 22:33

Add a ment  | 

4 Answers 4

Sorted by: Reset to default 5

Note that, as I mentioned in a ment above, you don't need to convert the value to a string; if you keep it as a decimal integer, you can use the bitshift and remainder operators to get your binary values.

var x = parseInt('A6', 16); // 166
var result = {};

for (var n = 1; x > 0; x = x >> 1, n = n * 2) {
   result[n] = x % 2 == 1;
}

console.log(JSON.stringify(result, undefined, 2));

You can do it without going to an intermediate binary string, but if you're just stuck on that final step you can do something like the following:

var bin = parseInt('0xa6', 16).toString(2);

Array.from(bin).reduce(function(byte, bit, index) {
    byte[1 << bin.length - index - 1] = bit === '1';
    return byte;
}, {});

JavaScript demo

Here's one long liner

const x = parseInt('A6', 16).toString(2).split('').reduceRight((last, bit, i, arr) => Object.assign(last, {[`${1 << (arr.length - i - 1)}`]: bit == 1}), {})
console.log(x)

It assumes that you're in an environment that supports puted keys (which is available in ES2015)

Here it is.

var x = {}, s = 1;
(0x10100110).toString(16).split('').reverse().map(a => {
    x[s] = !!~~a;
    s *= 2;
});

This is an odd one, but stick with me here. Given a hex value (from 00-FF), I need to convert it into binary, then from there convert it into an object with the key being the 'place' of each bit, and the value being a boolean.

That was an awkward explanation. Here's an example:

Given 'FF', I need to translate that into binary (parseInt(<hex value>, 16).toString(2) works for that job).

Now that FF has bee 1111 1111, I need to convert that into an object that looks like so:

{ 
  "1": true, 
  "2": true, 
  "4": true, 
  "8": true, 
  "16":true, 
  "32":true,
  "64": true, 
  "128": true
}

As another example, given the hex value 'A6', that bees 1010 0110 which would bee:

{ 
  "1": false, 
  "2": true, 
  "4": true, 
  "8": false, 
  "16":false, 
  "32":true, 
  "64": false, 
  "128": true
}

The only thing I'm looking to do is that final conversion from binary -> object.

I know I can do it with a simple loop, but I was wondering if there's any cool p-sci way to acplish this.

Thanks!

This is an odd one, but stick with me here. Given a hex value (from 00-FF), I need to convert it into binary, then from there convert it into an object with the key being the 'place' of each bit, and the value being a boolean.

That was an awkward explanation. Here's an example:

Given 'FF', I need to translate that into binary (parseInt(<hex value>, 16).toString(2) works for that job).

Now that FF has bee 1111 1111, I need to convert that into an object that looks like so:

{ 
  "1": true, 
  "2": true, 
  "4": true, 
  "8": true, 
  "16":true, 
  "32":true,
  "64": true, 
  "128": true
}

As another example, given the hex value 'A6', that bees 1010 0110 which would bee:

{ 
  "1": false, 
  "2": true, 
  "4": true, 
  "8": false, 
  "16":false, 
  "32":true, 
  "64": false, 
  "128": true
}

The only thing I'm looking to do is that final conversion from binary -> object.

I know I can do it with a simple loop, but I was wondering if there's any cool p-sci way to acplish this.

Thanks!

• javascript

Share Improve this question Follow asked May 31, 2016 at 22:11 Kieran EKieran E 3,67633 gold badges2121 silver badges4343 bronze badges 3

• 1 I can think of many cool more plicated ways to do it, but a loop seems in order in this case. Just don't calculate the numeric keys every time, do it once and store them. – Eran Goldin Commented May 31, 2016 at 22:16
• 2 Note that you don't need to convert the value into a string: x = parseInt('A6', 16) // 255; for (n = 1; x > 0; x = x >> 1, n = n * 2) { console.log(n, x % 2 == 1) } – Hamms Commented May 31, 2016 at 22:23
• @Hamms You should really post that (with a little extra) as an answer. It's trivial to go from that to the object OP wants to generate. – Mike Cluck Commented May 31, 2016 at 22:33

Add a ment  | 

4 Answers 4

Sorted by: Reset to default 5

Note that, as I mentioned in a ment above, you don't need to convert the value to a string; if you keep it as a decimal integer, you can use the bitshift and remainder operators to get your binary values.

var x = parseInt('A6', 16); // 166
var result = {};

for (var n = 1; x > 0; x = x >> 1, n = n * 2) {
   result[n] = x % 2 == 1;
}

console.log(JSON.stringify(result, undefined, 2));

You can do it without going to an intermediate binary string, but if you're just stuck on that final step you can do something like the following:

var bin = parseInt('0xa6', 16).toString(2);

Array.from(bin).reduce(function(byte, bit, index) {
    byte[1 << bin.length - index - 1] = bit === '1';
    return byte;
}, {});

JavaScript demo

Here's one long liner

const x = parseInt('A6', 16).toString(2).split('').reduceRight((last, bit, i, arr) => Object.assign(last, {[`${1 << (arr.length - i - 1)}`]: bit == 1}), {})
console.log(x)

It assumes that you're in an environment that supports puted keys (which is available in ES2015)

Here it is.

var x = {}, s = 1;
(0x10100110).toString(16).split('').reverse().map(a => {
    x[s] = !!~~a;
    s *= 2;
});

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