Replace array elements without losing reference?

Question

How does one replace all elements of an array without losing references?

var arr = [1, 2, 3];
var b = arr;
b == arr; // true
magic(arr, [4, 5, 6]);
b == arr; // should return true

One way of doing it is by popping and pushing. Is there a clean way?

Nina Scholz · Accepted Answer · 2017-06-20 18:34:22Z

10

You could splice the old values and append the new values.

function magic(reference, array) {
    [].splice.apply(reference, [0, reference.length].concat(array));
}

var arr = [1, 2, 3],
    b = arr;

console.log(b === arr); // true
magic(arr, [4, 5, 6]);
console.log(b === arr); // should return true

console.log(arr);

Another way, is to use Object.assign. This requires to set the length of the array, if it is smaller than the original array.

function magic(reference, array) {
    Object.assign(reference, array, { length: array.length });
}

var arr = [1, 2, 3],
    b = arr;

console.log(b === arr); // true
magic(arr, [4, 5, 6, 7]);
console.log(b === arr); // should return true

console.log(arr);

edited Jun 20, 2017 at 18:34

answered Jun 20, 2017 at 18:04

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

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5 Comments

Patrick Roberts Over a year ago

Nothing questionable, it just makes me smile

user663031 Over a year ago

@NinaScholz That seems like of a lot of work just to copy values from one array into another.

Scott Sauyet Over a year ago

@NinaScholz: What's the advantage of that [].apply over something like reference.splice(0, reference.length, ...array) ? Is it just es5 compatibility?

Nina Scholz Over a year ago

@ScottSauyet, it is just the compatibility with ES5. with ES6, you could use spread syntax or the above mentioned Object.assign. acually the request needs a mutator function.

Nina Scholz Over a year ago

@torazaburo, some action is still required while keeping the reference, either lenght and push, splice or Object.assign.

trincot · Accepted Answer · 2017-06-20 18:08:12Z

3

The magic part could be:

arr.splice(0, arr.length, 4, 5, 6);

var arr = [1, 2, 3];
var b = arr;
b == arr; // true
arr.splice(0, arr.length, 4, 5, 6);
console.log(b);
console.log(arr);
console.log(arr === b);

.as-console-wrapper { max-height: 100% !important; top: 0; }

If you already have the replacing array in a variable (let's say repl = [4, 5, 6]), then use the rest parameters syntax:

arr.splice(0, arr.length, ...repl);

var arr = [1, 2, 3];
var b = arr;
var repl = [4, 5, 6];
b == arr; // true

arr.splice(0, arr.length, ...repl);
console.log(b);
console.log(arr);
console.log(arr === b);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Jun 20, 2017 at 18:08

answered Jun 20, 2017 at 18:02

trincot

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Comments

Patrick Roberts · Accepted Answer · 2017-06-20 18:04:54Z

1

Here's one way:

var arr = [1, 2, 3];
var b = arr;
console.log(`b == arr, b
`, b == arr, b.join());
var c = magic(arr, [4, 5, 6]);
console.log(`b == arr, b
`, b == arr, b.join());
console.log(`c == arr, c
`, c == arr, c.join());

function magic(to, from) {
  // remove elements from existing array
  var old = to.splice(0);
  
  for (var i = 0; i < from.length; i++) {
    to[i] = from[i];
  }
  
  return old;
}

This implementation returns a copy of the old elements that were originally in the array.

answered Jun 20, 2017 at 18:04

Patrick Roberts

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Comments

user663031 · Accepted Answer · 2017-06-20 18:11:25Z

0

Copy the new values over the old ones.

function magic(arr, newvals) {
  for (let i = 0; i < newvals.length; i++) arr[i] = newvals[i];
  arr.length = newvals.length;
}

answered Jun 20, 2017 at 18:11

user663031

Comments

Gil Epshtain · Accepted Answer · 2019-09-23 16:45:13Z

0

function replaceArrValues(arrRef, newValues)
{
  arrRef.length = 0; // clear the array without losing reference
  newValues.forEach(x => arrRef.push(x));
}

answered Sep 23, 2019 at 16:45

Gil Epshtain

9,99110 gold badges78 silver badges100 bronze badges

2 Comments

Wyck Over a year ago

This code-only answer could be improved with a little bit of written explanation. Especially: how does this compare to the splice technique others have contributed.

nircraft Over a year ago

While this code snippet may be the solution, including an explanation really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion.

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