Javascript return array of objects [duplicate]

Question

function setup() {
  var names = [];
  var name = {firstname: "", lastname: ""};

  name.firstname = "John";
  name.lastname = "Doe";
  names.push(name);

  name.firstname = "Bill";
  name.lastname = "Smith";
  names.push(name);

  return names;
}

var temp = setup();
print temp[0].firstname;

I can't seem to figure out how to return an array of objects from a function. Any idea where I'm going wrong?

The problem is that the result stored in temp is the following:

 [
   {
     firstname: "Bill",
     lastname: "Smith"
   },
   {
     firstname: "Bill",
     lastname: "Smith"
   }
 ]

You are returning an array... that's why temp[0] works. What's the issue? — ryan0319
– ryan0319, Commented Jan 19, 2016 at 20:33
you are pushing a reference to the same object into the names array, and overwriting the properties of that object. it will mean you have duplicate entries of bill smith in the output array. — synthet1c
– synthet1c, Commented Jan 19, 2016 at 20:36
FYI, aside from the fact that JS doesn't deep copy objects automatically, you can actually shorten your code a good bit by simply using a literal structure. No need for .push(). var names = [{firstname: "john", lastname: "doe"}, {firstname: "bill": lastname: "smith"}]; You can also pass multiple items to .push() at once. — user1106925
– user1106925, Commented Jan 19, 2016 at 20:44

Robert McKee · Accepted Answer · 2016-01-19 20:49:41Z

6

You can't reuse name to push the "second" object because you haven't created a second object, it is the same object. So name.firstname="Bill" is changing the first object not creating a second one.

function setup() {
  var names = [];

  var name = {};
  name.firstname = "John";
  name.lastname = "Doe";
  names.push(name);

  var name2 = {};
  name2.firstname = "Bill";
  name2.lastname = "Smith";
  names.push(name2);

  return names;
}

var temp = setup();
alert(temp[0].firstname); // Alerts "John"

edited Jan 19, 2016 at 20:49

answered Jan 19, 2016 at 20:34

Robert McKee

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Comments

Nikhil Aggarwal · Accepted Answer · 2016-01-19 20:48:17Z

5

There are 2 problems with your code.

To paint in console, you need to use console.log(temp[0].firstname) in place of print
You have a name object and you are updating it only. And as objects are passed by reference, both the instances at index 0 and index 1 are updated and have same value.

You need to update your code to following

function setup() {
      var names = [];
      var n1 = {};
    
      n1.firstname = "John";
      n1.lastname = "Doe";
      names.push(n1);
    
      var n2 = {};
      n2.firstname = "Bill";
      n2.lastname = "Smith";
      names.push(n2);
    
      return names;
    }
    
    var temp = setup();
    console.log(temp[0].firstname);

edited Jan 19, 2016 at 20:48

answered Jan 19, 2016 at 20:38

Nikhil Aggarwal

28.5k4 gold badges46 silver badges61 bronze badges

8 Comments

Nikhil Aggarwal Over a year ago

@Downvoter - reason please

Mike Cluck Over a year ago

No idea what the downvoters problem was. This is a reasonable answer.

Nikhil Aggarwal Over a year ago

@NikolayErmakov - I was holding back while posting this. But at the time I posted my answer Robert Mckee's answer did not had any explanation. And I wanted to add that. Whats wrong with that

Robert McKee Over a year ago

Nothing wrong with this answer, and I've upvoted it as it is actually worded clearer than my own.

Nikolay Ermakov Over a year ago

I upvoted the answer back

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