Kevin Powell - Community•3mo ago

Array().fill acts weirdly

// Option 1
    function initialize() {
        let matrix = new Array(3).fill(["-", "-", "-"]);
        return matrix;
    };
// Option 2
    function initialize() {
        let matrix = new Array();
        for (; matrix.length < 3;) {
            let row = ["-", "-", "-"];
            matrix.push(row);
        };
        return matrix;
    };

// Option 1
    function initialize() {
        let matrix = new Array(3).fill(["-", "-", "-"]);
        return matrix;
    };
// Option 2
    function initialize() {
        let matrix = new Array();
        for (; matrix.length < 3;) {
            let row = ["-", "-", "-"];
            matrix.push(row);
        };
        return matrix;
    };

If I do matrix[0][0] = "a" in option 1 matrix, it replaces entire 1st column with "a". But that's not the case for option 2, it only only replaces at the specified position.

5 Replies

ἔρως•3mo ago

because you're filling the array with the same reference to the array you're not filling with 3 different copies of the same array this is the same as doing this:

let x = [1, 2, 3];
let y = [x, x, x];

let x = [1, 2, 3];
let y = [x, x, x];

oh, and the 2nd one is the equivalent of this:

let x = [
  [1, 2, 3],
  [1, 2, 3],
  [1, 2, 3]
];

let x = [
  [1, 2, 3],
  [1, 2, 3],
  [1, 2, 3]
];

so, when you change the index on the first, you're actually changing the index on the same array for all

brrOP•3mo ago

with the same reference

brrOP•3mo ago

https://tenor.com/bnHxa.gif

Tenor

brrOP•3mo ago

thanks

ἔρως•3mo ago

you're welcome

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Array().fill acts weirdly

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