Please refer - /

var o = {
    x: 6
};

function a() {
    this.x = 3;
    this.y = function() {
        alert(this.x)
    }
}

var a1 = new a();
a1.y.call(); //shouldn't it alert 3?

a1.y.call(o) //is alerting 6, correct!

Why is the first alert, alerting undefined? shouldn't it alert 3?

Please refer - https://jsfiddle/ta2u2z9a/

var o = {
    x: 6
};

function a() {
    this.x = 3;
    this.y = function() {
        alert(this.x)
    }
}

var a1 = new a();
a1.y.call(); //shouldn't it alert 3?

a1.y.call(o) //is alerting 6, correct!

Why is the first alert, alerting undefined? shouldn't it alert 3?

• javascript
• call

Share Improve this question Follow edited Jul 8, 2015 at 14:04 jcubic 66.8k5858 gold badges249249 silver badges455455 bronze badges asked Jul 8, 2015 at 14:03 anand patilanand patil 50722 gold badges99 silver badges2727 bronze badges 2

• Change a1.y.call() => a1.y.call(a1); – elad.chen Commented Jul 8, 2015 at 14:05
• 3 Why would you do this? – Evan Davis Commented Jul 8, 2015 at 14:05

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2 Answers 2

Sorted by: Reset to default 10

The first argument to .call() needs to be the "owner object" of the method:

a1.y.call(a1);

What's going on?

When you call a method "naturally" (e.g. a1.y()), the value of this is automatically set to the object (e.g. a1).

When using the function not as a method (e.g. var x = a1.y; x();), the value of this is either the global/window object, or null (depending on whether you're in "strict mode" or not).

The .call() method (it's a method of the function) is a way to explicitly set the this value, instead of having it automatically assigned. When using .call(), you have to supply the this value yourself, as the first argument.

Because you have not sent the expected "this" parameter in call() as an argument so this is undefined in that function.

If you are not passing any argument to call() then no need to use it. Simply call the method on the object.