Why do these two javascript 2d-arrays behave differently? - Stack Overflow

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In my function, I have defined two arrays, the first (array1), has a pre-initialized length. I added the second array (array2) just for testing because I thought the first was behaving strangely.

My code:

function test(n = 3) {
  array1 = new Array(n).fill(new Array(n));
  array2 = [
    [undefined, undefined, undefined],
    [undefined, undefined, undefined],
    [undefined, undefined, undefined]
  ];

  document.getElementById("output").innerHTML = JSON.stringify(array1) + " (array 1) <br/>" + JSON.stringify(array2) + " (array 2)<br/><br/><hr/>";


  for (i = 0; i < n; i++) {
    array1[i][0] = i;
    array2[i][0] = i;
  }

  document.getElementById("output").innerHTML += JSON.stringify(array1) + " (array 1) <br/>" + JSON.stringify(array2) + " (array 2)<br/><br/><hr/>";

}

<button onclick="test();">Press to test</button>

<br/><br/>
<div id="output"></div>

In my function, I have defined two arrays, the first (array1), has a pre-initialized length. I added the second array (array2) just for testing because I thought the first was behaving strangely.

My code:

function test(n = 3) {
  array1 = new Array(n).fill(new Array(n));
  array2 = [
    [undefined, undefined, undefined],
    [undefined, undefined, undefined],
    [undefined, undefined, undefined]
  ];

  document.getElementById("output").innerHTML = JSON.stringify(array1) + " (array 1) <br/>" + JSON.stringify(array2) + " (array 2)<br/><br/><hr/>";


  for (i = 0; i < n; i++) {
    array1[i][0] = i;
    array2[i][0] = i;
  }

  document.getElementById("output").innerHTML += JSON.stringify(array1) + " (array 1) <br/>" + JSON.stringify(array2) + " (array 2)<br/><br/><hr/>";

}

<button onclick="test();">Press to test</button>

<br/><br/>
<div id="output"></div>

In the for loop, I try to change the first value of the second dimensions. It should output [[0, undefined, undefined], [1, undefined, undefined], [2, undefined, undefined]], like the second array does.

My questions are: why does this happen? And, how can I make a pre-initialized array with length n in both dimensions, that behaves like the second array?

• javascript
• arrays
• multidimensional-array

Share Improve this question Follow edited Aug 20, 2017 at 14:36 mplungjan 179k2828 gold badges182182 silver badges240240 bronze badges asked Aug 20, 2017 at 14:29 SachaSacha 8931010 silver badges2929 bronze badges 1

• Sharing an article, why not to use new Array(), hope this helps -> coderwall./p/h4xm0w/why-never-use-new-array-in-javascript – mindaJalaj Commented Aug 20, 2017 at 14:38

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 9

That's because in case of array1 it contains three arrays, but all three of them point to the same reference variable, that was evaluated when new Array(n) was executed:

var array1 = new Array(n).fill(new Array(n));

So when the for loop runs over array1 it is setting the value of the same array reference, while in case of array2 those three arrays are different reference variables.

Here's a slightly modified version of your snippet. Notice the entries into console when the value of array1's element is being changed. In case of array1 all three child arrays are changing, while in case of array2 the array referenced under the loop using index i is the only one that changes.

function test(n = 3) {
  array1 = new Array(n).fill(new Array(n));
  array2 = [
    [undefined, undefined, undefined],
    [undefined, undefined, undefined],
    [undefined, undefined, undefined]
  ];

  document.getElementById("output").innerHTML = JSON.stringify(array1) + " (array 1) <br/>" + JSON.stringify(array2) + " (array 2)<br/><br/><i>The mas with nothing in between mean undefined.</i><hr/>";


  for (i = 0; i < n; i++) {
    array1[i][0] = i;
    array2[i][0] = i;
    console.log("Array 1: " + JSON.stringify(array1));
    console.log("Array 2: " + JSON.stringify(array2));
  }

  document.getElementById("output").innerHTML += JSON.stringify(array1) + " (array 1) <br/>" + JSON.stringify(array2) + " (array 2)<br/><br/><i>The mas with nothing in between mean undefined.</i><hr/>";

}

<button onclick="test();">Press to test</button>

<br/><br/>
<div id="output"></div>

Because Array.fill

The fill() method fills all the elements of an array from a start index to an end index with a static value.

takes a static value and fills the array with it. Therefore you get in every element of array1 the same array of the filling.

function test(n = 3) {
    var array1 = new Array(n).fill(new Array(n)),
        array2 = [[undefined, undefined, undefined], [undefined, undefined, undefined], [undefined, undefined, undefined]],
        i;

    for (i = 0; i < n; i++) {
        array1[i][0] = i;
        array2[i][0] = i;
    }

    document.getElementById("output").innerHTML = array1 + " (array 1) <br/>" + array2 + " (array 2)<br/><br/><i>The mas with nothing in between mean undefined.</i>";
    console.log(array1);
    console.log(array2);
}

<button onclick="test();">Press to test</button><br/><br/>
<div id="output"></div>

To get an independent filled array, you could use Array.from and map a new array with mapped values.

var array = Array.from({ length: 3 }, _ => Array.from({ length: 3 }, _ => 4));

array[0][0] = 0;
console.log(array);

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

In my function, I have defined two arrays, the first (array1), has a pre-initialized length. I added the second array (array2) just for testing because I thought the first was behaving strangely.

My code:

function test(n = 3) {
  array1 = new Array(n).fill(new Array(n));
  array2 = [
    [undefined, undefined, undefined],
    [undefined, undefined, undefined],
    [undefined, undefined, undefined]
  ];

  document.getElementById("output").innerHTML = JSON.stringify(array1) + " (array 1) <br/>" + JSON.stringify(array2) + " (array 2)<br/><br/><hr/>";


  for (i = 0; i < n; i++) {
    array1[i][0] = i;
    array2[i][0] = i;
  }

  document.getElementById("output").innerHTML += JSON.stringify(array1) + " (array 1) <br/>" + JSON.stringify(array2) + " (array 2)<br/><br/><hr/>";

}

<button onclick="test();">Press to test</button>

<br/><br/>
<div id="output"></div>

In my function, I have defined two arrays, the first (array1), has a pre-initialized length. I added the second array (array2) just for testing because I thought the first was behaving strangely.

My code:

function test(n = 3) {
  array1 = new Array(n).fill(new Array(n));
  array2 = [
    [undefined, undefined, undefined],
    [undefined, undefined, undefined],
    [undefined, undefined, undefined]
  ];

  document.getElementById("output").innerHTML = JSON.stringify(array1) + " (array 1) <br/>" + JSON.stringify(array2) + " (array 2)<br/><br/><hr/>";


  for (i = 0; i < n; i++) {
    array1[i][0] = i;
    array2[i][0] = i;
  }

  document.getElementById("output").innerHTML += JSON.stringify(array1) + " (array 1) <br/>" + JSON.stringify(array2) + " (array 2)<br/><br/><hr/>";

}

<button onclick="test();">Press to test</button>

<br/><br/>
<div id="output"></div>

In the for loop, I try to change the first value of the second dimensions. It should output [[0, undefined, undefined], [1, undefined, undefined], [2, undefined, undefined]], like the second array does.

My questions are: why does this happen? And, how can I make a pre-initialized array with length n in both dimensions, that behaves like the second array?

• javascript
• arrays
• multidimensional-array

Share Improve this question Follow edited Aug 20, 2017 at 14:36 mplungjan 179k2828 gold badges182182 silver badges240240 bronze badges asked Aug 20, 2017 at 14:29 SachaSacha 8931010 silver badges2929 bronze badges 1

• Sharing an article, why not to use new Array(), hope this helps -> coderwall./p/h4xm0w/why-never-use-new-array-in-javascript – mindaJalaj Commented Aug 20, 2017 at 14:38

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 9

That's because in case of array1 it contains three arrays, but all three of them point to the same reference variable, that was evaluated when new Array(n) was executed:

var array1 = new Array(n).fill(new Array(n));

So when the for loop runs over array1 it is setting the value of the same array reference, while in case of array2 those three arrays are different reference variables.

Here's a slightly modified version of your snippet. Notice the entries into console when the value of array1's element is being changed. In case of array1 all three child arrays are changing, while in case of array2 the array referenced under the loop using index i is the only one that changes.

function test(n = 3) {
  array1 = new Array(n).fill(new Array(n));
  array2 = [
    [undefined, undefined, undefined],
    [undefined, undefined, undefined],
    [undefined, undefined, undefined]
  ];

  document.getElementById("output").innerHTML = JSON.stringify(array1) + " (array 1) <br/>" + JSON.stringify(array2) + " (array 2)<br/><br/><i>The mas with nothing in between mean undefined.</i><hr/>";


  for (i = 0; i < n; i++) {
    array1[i][0] = i;
    array2[i][0] = i;
    console.log("Array 1: " + JSON.stringify(array1));
    console.log("Array 2: " + JSON.stringify(array2));
  }

  document.getElementById("output").innerHTML += JSON.stringify(array1) + " (array 1) <br/>" + JSON.stringify(array2) + " (array 2)<br/><br/><i>The mas with nothing in between mean undefined.</i><hr/>";

}

<button onclick="test();">Press to test</button>

<br/><br/>
<div id="output"></div>

Because Array.fill

The fill() method fills all the elements of an array from a start index to an end index with a static value.

takes a static value and fills the array with it. Therefore you get in every element of array1 the same array of the filling.

function test(n = 3) {
    var array1 = new Array(n).fill(new Array(n)),
        array2 = [[undefined, undefined, undefined], [undefined, undefined, undefined], [undefined, undefined, undefined]],
        i;

    for (i = 0; i < n; i++) {
        array1[i][0] = i;
        array2[i][0] = i;
    }

    document.getElementById("output").innerHTML = array1 + " (array 1) <br/>" + array2 + " (array 2)<br/><br/><i>The mas with nothing in between mean undefined.</i>";
    console.log(array1);
    console.log(array2);
}

<button onclick="test();">Press to test</button><br/><br/>
<div id="output"></div>

To get an independent filled array, you could use Array.from and map a new array with mapped values.

var array = Array.from({ length: 3 }, _ => Array.from({ length: 3 }, _ => 4));

array[0][0] = 0;
console.log(array);

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

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