javascript: unexpected behavior pushing into empty array

Question

The following code:

var arr1 = [1,2,3];
var obj1 = {};

for (var j = 0; j < arr1.length; j++) {
    if (obj1[j.toString()])
        obj1[j.toString()] = obj1[j.toString()].push(j)
    else
        obj1[j.toString()] = [].push(j);
}

produced the following output:

   obj1
=> { '0': 1, '1': 1, '2': 1 }

and I would just kindly like to know why.

(I'm aware now that the following code:

var arr1 = [1,2,3];
var obj1 = {};

for (var j = 0; j < arr1.length; j++) {
    if (obj1[j.toString()])
        obj1[j.toString()] = obj1[j.toString()].push(j)
    else {
        obj1[j.toString()] = [];
        obj1[j.toString()].push(j);
    }
}

will give me my desired output:

   obj1
=> { '0': [ 0 ], '1': [ 1 ], '2': [ 2 ] }

)

Well, [] is the same as Array.prototype, there is no empty array, and you're not pushing to anything inside obj. The second code snippet is the correct way of doing it, where you actually create an empty array — adeneo
– adeneo, Commented Mar 2, 2016 at 19:48

cl3m · Accepted Answer · 2021-06-16 07:41:42Z

11

Because, as per the documentation, the Array.prototype.push() method returns the Array length, not the array itself.

You may prefer the concat method like so:

var arr1 = [1,2,3];
var obj2 = {}

for (var j = 0; j < arr1.length; j++) {
  var js = j.toString()
  if (obj2[js]) {
    obj2[js] = obj2[js].concat([j])
  } else {
    obj2[js] = [j]
  }
}
console.log(obj2) // => { '0': [ 0 ], '1': [ 1 ], '2': [ 2 ] }

// shorter version
var obj3 = {}

for (var j = 0; j < arr1.length; j++) {
  var js = j.toString()
  obj3[js] = [].concat(obj3[js] || [], [j])
}

console.log(obj3) // => { '0': [ 0 ], '1': [ 1 ], '2': [ 2 ] }

edited Jun 16, 2021 at 7:41

answered Mar 2, 2016 at 19:47

cl3m

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2 Comments

Rajaprabhu Aravindasamy Over a year ago

Did not notice the first snippet. +1

anatta Over a year ago

Confirmed. [1,2,3].push(60); // 4; [].push(90); // 1; Thanks.

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