Unexpected value of array in nested loop

Question

while practicing with loops I receive kind of weird to me result.

let a = [];
let b = [];

for (let i=0; i<2; i++){
  for (let j=0; j<2;j++){
      a.push(0);

  }
  b.push(a);
  console.log(a);
  console.log(b);
}

Output looks like this:

//values of arrays after first iteration of outside loop
[ 0, 0 ]
[ [ 0, 0 ] ]
//values of arrays after second...
[ 0, 0, 0, 0 ]
[ [ 0, 0, 0, 0 ], [ 0, 0, 0, 0 ] ]

I wonder why value of array b is not equal to [ [0, 0], [0, 0, 0, 0] ]?

You never make a copy of a, so the same array is pushed onto b each time through the outer loop. — Pointy
– Pointy, Commented Dec 5, 2018 at 17:01

dquijada · Accepted Answer · 2018-12-05 17:23:05Z

1

Basically you are always using the same a array.

At the start of the second iteration a is [0, 0] and you push 2 extra 0s there.

And b is basically [a, a] in the end, so if you keep feeding a 0s, it's elements will keep growing.

If you want to create a 2D array you would have to make this little modification:

let b = [];

for (let i=0; i<2; i++){
  let a = [];
  for (let j=0; j<2;j++){
      a.push(0);

  }
  b.push(a);
  console.log(a);
  console.log(b);
}

Moving the creation of let a into the outer loop means every time that cycles you will create a new, temporal a to be later pushed to b.

To get the result you want you need to make a copy of the array a, instead, making it:

let a = [];
let b = [];

for (let i=0; i<2; i++){
  for (let j=0; j<2;j++){
      a.push(0);
}
  b.push(a.slice());
  console.log(a);
  console.log(b);
}

edited Dec 5, 2018 at 17:23

answered Dec 5, 2018 at 17:09

dquijada

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4 Comments

Keith Over a year ago

That's not going to give the result the OP is after.

dquijada Over a year ago

Unless I misunderstood what he wanted, it will. Could you elaborate?

Keith Over a year ago

In his question he expected b = [ [0, 0], [0, 0, 0, 0] ] But with this code it's going to be [[0,0],[0,0]] instead,. The OP has accepted your answer so, maybe he does want [[0,0],[0,0]]..

dquijada Over a year ago

Oh, true, I indeed misunderstood. Gonna edit it right away

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