Fail to set matrix cell value

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Why when I do this:

let matrix = Array(3).fill(new Array(3).fill(0));
matrix[0][1] = 1

I would expect the matrix to be:

    [ [0,1,0],
      [0,0,0]  ]

But instead, the matrix looks like this:

   [ [0,1,0],
     [0,1,0] ]

So it's setting the entire column rather than just the cell. I feel I'm missing a very important piece of the language here.

Thank you in advance!

Because you're using the same reference in memory for all indexes.

Imagine this is the memory, the indexes are pointing to the same value in memory, therefore, every change in a specific index will modify that value

    new Array(3).fill(0)     Matrix
    +---------+           +-----------+
    | [0,0,0] | <--+------|   Index 0 |
    +---------+    |      +-----------+
          ^        +------|   Index 1 |
          |               +-----------+
          +---------------|   Index 2 |

Create a new array for each index

    {length: 3}, () => Array(3).fill(0)     Matrix
    +---------+                         +-----------+
    | [0,0,0] | <-----------------------|   Index 0 |
    +---------+                         +-----------+
    | [0,0,0] | <-----------------------|   Index 1 |
    +---------+                         +-----------+
    | [0,0,0] | <-----------------------|   Index 2 |
    +---------+                         +-----------+

let matrix = Array.from({length: 3}, () => Array(3).fill(0));
matrix[0][1] = 1;


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When you use .fill and pass it a non-primitive (like an object or array), there will only be one of those actual objects in memory; the array will be filled with three references to the same object, so when one index gets mutated, all of them do. If you want to fill the array with separate objects, you'll have to explicitly create them on each iteration of creating the new array, which you might do with Array.from:

const matrix = Array.from(
  { length: 3 },
  () => new Array(3).fill(0)
matrix[0][1] = 1;

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Since the "outer" array indices are only referencing one array you can also initially fill the initial array with zeros and then use map to create new arrays at each index:

let matrix = Array(3).fill(0).map(n => new Array(3).fill(0));

matrix[0][1] = 1;

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