Reputation: 861
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?
Upvotes: 6
Views: 182
Reputation: 386570
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 commas 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; }
Upvotes: 7
Reputation: 14541
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 commas 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 commas with nothing in between mean undefined.</i><hr/>";
}
<button onclick="test();">Press to test</button>
<br/><br/>
<div id="output"></div>
Upvotes: 9