JavaScriptのループとスコープ

ループとスコープに関連して、開発者にいくつかの頭痛の種を引き起こす可能性のあるJavaScriptの機能が1つあります。ループとvarを使用したスコープに関するいくつかのトリックを学び、

の1つの機能がありますJavaScriptこれは、ループやスコープに関連して、開発者にいくつかの頭痛の種を引き起こす可能性があります。

この例を見てください:

const operations = []

for (var i = 0; i < 5; i++) { operations.push(() => { console.log(i) }) }

for (const operation of operations) { operation() }

基本的に反復し、5回、operationsという関数を配列に追加します。この関数コンソールは、ループインデックス変数をログに記録しますi

後でこれらの関数を実行します。

ここで期待される結果は次のようになります。

0
1
2
3
4

but actually what happens is this:

5
5
5
5
5

Why is this the case? Because of the use of var.

Since var declarations are hoisted, the above code equals to

var i;
const operations = []

for (i = 0; i < 5; i++) { operations.push(() => { console.log(i) }) }

for (const operation of operations) { operation() }

so, in the for-of loop, i is still visible, it’s equal to 5 and every reference to i in the function is going to use this value.

So how should we do to make things work as we want?

The simplest solution is to use let declarations. Introduced in ES6, they are a great help in avoiding some of the weird things about var declarations.

Changing var to let in the loop variable is going to work fine:

const operations = []

for (let i = 0; i < 5; i++) { operations.push(() => { console.log(i) }) }

for (const operation of operations) { operation() }

Here’s the output:

0
1
2
3
4

How is this possible? This works because on every loop iteration i is created as a new variable each time, and every function added to the operations array gets its own copy of i.

Keep in mind you cannot use const in this case, because there would be an error as for tries to assign a new value in the second iteration.

Another way to solve this problem was very common in pre-ES6 code, and it is called Immediately Invoked Function Expression (IIFE).

In this case you can wrap the entire function and bind i to it. Since in this way you’re creating a function that immediately executes, you return a new function from it, so we can execute it later:

const operations = []

for (var i = 0; i < 5; i++) { operations.push(((j) => { return () => console.log(j) })(i)) }

for (const operation of operations) { operation() }


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