Understanding console.log(console.log(object))

Question

I am new to programming and js and I am trying to learn crux of javascript.

var obj1 = {
    name: 'rawn',
    fn: function() {
        console.log(this);
    }
};
console.log(obj1.fn());

When I output this I get the object(as expected) - {name: "rawn", fn: ƒ} and on another line I get - undefined. So my question is why and how do I get undefined?

My understanding is that, we are writing this line - console.log(obj1.fn()); as , console.log(console.log(this)), so how does javascript engine gives the result as undefined (what was put as undefined in the execution context)?

What would you expect as result here…?

deceze
– deceze ♦

2018-11-10 10:50:42 +00:00
Commented Nov 10, 2018 at 10:50 — deceze
– deceze ♦, Commented Nov 10, 2018 at 10:50
@deceze i though it will simply print the object 2 times

Rohan Sharma
– Rohan Sharma

2018-11-10 11:29:40 +00:00
Commented Nov 10, 2018 at 11:29 — Rohan Sharma
– Rohan Sharma, Commented Nov 10, 2018 at 11:29

deceze · Accepted Answer · 2018-11-10 10:49:54Z

2

console.log(obj1.fn()) and console.log(console.log(this)) are not equivalent at all, but they actually have the same result for the same reason: the inner function does not return anything. The return value of these functions is undefined. Which is what the outer console.log logs.

answered Nov 10, 2018 at 10:49

deceze♦

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3 Comments

Rohan Sharma Over a year ago

console.log(obj1.fn()) and console.log(console.log(this)) are not equivalent at all, but I thought when I say () in the fn(), it executes the my written code part i.e console.log(this)

Rohan Sharma Over a year ago

oh yeah, i din't mean it literally, I meant console.log(this) is called first from the obj.fn() and then the outer console.log is called

Rohan Sharma Over a year ago

If the function is passed to the function console.log(), then the function will display the value of the passed function().

Now I know whats going on !

Sid · Accepted Answer · 2018-11-10 10:49:26Z

1

This is because of two reasons. First, in console.log(this); the context of this inside the fn function is the object itself. Hence it prints the guts of obj1.

Secondly, in console.log(obj1.fn());, you are actually calling the method which returns nothing. Instead if you say console.log(obj1.fn); it will return:

ƒ () { console.log(this); }

Hope this helps.

answered Nov 10, 2018 at 10:49

Sid

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