Understanding the behavior of JavaScript call function

Question

Let I have a simple code

{
  let i = 10;
  function outer() {
    let j = 20;
    console.log(i, j);
  }
  let inner = outer();
}

And the output is

10 20

which is obvious. But I thought I only stored the function value in inner, but never called it. Then why is it logging the output to the console?

Not sure what exactly you want to know. You call the function right there in your code. inner is equal to the return value of your outer function: undefined. If you want to copy the reference to your function, you’d use let inner = outer; instead. — Sebastian Simon
– Sebastian Simon, Commented Jul 20, 2020 at 5:05
Thank you for your clarification. I almost forgot that bracket part. — isuro
– isuro, Commented Jul 20, 2020 at 5:07

mgm793 · Accepted Answer · 2020-07-20 05:08:15Z

1

let inner = outer() => executes the function

let inner = outer => save the function to inner

As you can see the difference is the parentheses.

When you use parentheses, you are saying execute this function.

answered Jul 20, 2020 at 5:08

mgm793

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Comments

Enrigle Robota · Accepted Answer · 2020-07-20 07:09:05Z

1

The outmost braces just indicate they make a "code block", not a function body.

In outer function, it initializes a local variable 'j' and refer to a external variable i. And you called the function with let inner = outer().

answered Jul 20, 2020 at 7:09

Enrigle Robota

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