Closure..how it works..

Question

How it works...
```javaSCript
function makeAdder(x) {
  return function(y) {
    return x + y;
  };
}
var add5 = makeAdder(5);
var add10 = makeAdder(10);
console.log(add5(2));  // 7
console.log(add10(2)); // 12
```
This is my understanding..  
1) base on var add5 = makeAdder(5); If put 5 to function makeAdder(x) I get 5 + y, what is the value of y ? undefined ?   
2) console.log(add5(2)); x is already exist 5, how can I put the value 2 and get the answer of 7?

Kris Byrum · Accepted Answer

y hasn't been declared yet so it's nothing at the moment.      
When you call this type of closure what you're doing is setting the local scope. The local scope in this instance will set x to 5. You don't have a way to change that so x will always be 5.
The return of the makeAdder function will be another function. So whatever variable you assign makeAdder to, will now evaluate to that return function. 
If you want to get 7, then you simply pass in 2 to the variable you make for makeAdder. (keep in mind you can call it like makeAdder(x)(y) and it will do the same. No need to assign it to a variable)
eg.
```
function closure(x) {
    const localScopeVar = x
    return function(y) {
        return localScopeVar + y
    }
}
const firstCall = closure(5) // evaluates to function(y) { return 5 + y }
const secondCall = firstCall(2) // evaluates to 7
```

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Tani Huang

Tani Huang

Closure..how it works..

1 Answer

Kris Byrum

Kris Byrum