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javascriptclosures
kazinix
kazinix

Reputation: 30163

Does a parameter variable create closure in JavaScript?

I am tasked to trace the cause of a memory leak in one of our applications so I'm trying to study closures. I wonder if this code creates closure:

function foo(p)
{
    return function(){ return p + 1; }
}

Based on my understanding, closure is created when the inner function gains access to a local variable of its parent function. The parameter p is local to foo, if the inner function gains an access to p, does it mean a closure is created?

Upvotes: 1

Views: 65

Answers (3)

jhamm
jhamm

Reputation: 25062

This does create a closure, but it is not the typical closure that is talke about in javascript. What is the typical example is:

var adder = function(a) {
  return function(b) {
    return a + b;
  }
}

What this does is give you the ability to create a "closure" or close in a variable to be used over and over again. I can create a function:

var adder4 = adder(4);

Now if I want to add 4 to any number I can user adder4(2) and the result in this case would be 6. What is going on here is that the 4 is inserted for the variable a. The variable a is then enclosed inside this function permanently. We can then replace the variable b at any time to create a new function. So when I made the function call adder4(2), I am using a function in which a has already been assigned. The variable 2 in that case is assigned to the variable b. Obviously when you add 4 and 2 you get 6. But you can use the same function to add another number, adder4(3). Now this does the same logic and gives you 7. The 4 is still enclosed, or in the "closure", but you are free to replace another variable in the middle of the function.

You also see this a lot with anonymous functions and click handlers, but you can google that for much better answers.

Hope this helps.

Upvotes: 0

Arun P Johny
Arun P Johny

Reputation: 388446

The parameters of a function exists in the local scope of the function, so yes it creates a closure

Upvotes: 1

Marius Schulz
Marius Schulz

Reputation: 16470

Yes, that is exactly what's happening here. The inner function you're returning has access to the parameter p through the local scope, so you're correct.

It would also have created a closure if you referenced a local variable from the outer function in the returning function, like that:

function foo(p) {
    var q = 4;
    return function() { return p + q; }
}

Here's a very detailed explanation: Explaining JavaScript Scope And Closures

Upvotes: 1

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