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],
      [0,0,0]  ]

But instead, the matrix looks like this:

   [ [0,1,0],
     [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;

console.log(matrix);

Assign value zero to n elements of cell array, I am declaring a cell array to hold some numeric and non numeric data. I want to set the first row, all columns, equal to the numeric value zero. Something like  I would like to reference the value of a cell in a RDLC matrix in order to set a property for other cells in the matrix body. More specifically, I would like to set the entire row's background color based on the value of the cell in the first column.


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;
console.log(matrix);

Assign empty value(s) to matrix elements (to delete those elements , But this fails when I don't directly specify the empty value, but use an empty the last iteration just removes the last column of the result matrix. Test a value against a one dimensional array and return either 1 or 0 depending on if there is an equal value in the array. I want to test a value against a range of values contained in a one dimensional array.


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;
console.log(matrix);

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