5 Pitfalls About JS Scope (Summary)

In JavaScript, a block, function, or module creates a scope for a variable. For example, the if code block creates a scope for the variable message:

if (true) {
  const message = 'Hello';
  console.log(message); // 'Hello'
}
console.log(message); // throws ReferenceError

message## can be accessed within the scope of the if code block #. But outside the scope, the variable is not accessible.

Okay, that’s a brief introduction to scopes. If you want to learn more, I recommend reading my article

JavaScript Scope Explained in Simple Words.

Here are 5 interesting situations where JavaScript scope behaves differently than you expect. You might study these cases to improve your understanding of scope, or simply to prepare for an interview.

1. for var Variables inside the loop

Consider the following code snippet:

const colors = ['red', 'blue', 'white'];

for (let i = 0, var l = colors.length; i < l; i++) {
  console.log(colors[i]); // &#39;red&#39;, &#39;blue&#39;, &#39;white&#39;
}
console.log(l); // ???
console.log(i); // ???

What happens when you print the

l and i variables?

Answer

##console.log(l)

outputs the number 3, while console.log(i ) throws ReferenceError.

l

Variables are declared using the var statement. As you may already know, var variables are only limited by the scope of the function body and not the code block. In contrast, variables

i

are declared using the let statement. Because the let variable is block-scoped, i is only accessible within the for loop scope.

Fix

Change the

l

declaration from var l = colors.length to const l = colors .length. Now the variable l is encapsulated in the for loop body.

2. Function declaration in code block

In the following code segment:

// ES2015 env
{
  function hello() {
    return 'Hello!';
  }
}

hello(); // ???

Call

hello()

will how? (The code snippet is executed in the ES2015 environment)

Answer

Because the code block creates a scope for the function declaration, it is executed in the ES2015 environment Calling

hello()

will cause ReferenceError: hello is not defined. Interestingly, in a pre-ES2015 environment, no error is thrown when executing the above code snippet.

Do you know why? Please write your answers in the comments below!

3. Where can you import modules?

Can you import modules in code blocks?

if (true) {
  import { myFunc } from 'myModule'; // ???
  myFunc();
}

Answer

The above script will trigger the error:

'import' and 'export' may only appear at the top-level

. You can only import modules in the top-level scope of the module file (also called

module scope

).

Fix

Always import modules from module scope. Another good practice is to place the

import

statement at the beginning of the source file: <pre class='brush:php;toolbar:false;'>import { myFunc } from &amp;#39;myModule&amp;#39;; if (true) { myFunc(); }</pre>ES2015’s module system is static. Determine module dependencies by analyzing JavaScript source code rather than executing code. So

import

statements cannot be included in code blocks or functions because they are executed at runtime.

4. Function parameter scope

Think about the following function:

let p = 1;

function myFunc(p = p + 1) {
  return p;
}

myFunc(); // ???

What happens when you call

myFunc()

?

Answer

When calling function

myFunc()

, an error will be thrown: ReferenceError: Cannot access 'p' before initialization . This happens because the parameters of the function have their own scope (separate from the function scope). The argument

p = p 1

is equivalent to let p = p 1. Let's take a closer look at

p = p 1

. First, define the variable

p

. JavaScript then tries to evaluate the default value expression p 1, but at this point the binding p has been created but not yet initialized (cannot access variables from the outer scope let p = 1). Therefore an error is thrown that p was accessed before initialization.

Fix

To fix this problem, you can rename the variable

let p = 1

and you can also rename the function parameter p = p 1. Let's choose to rename the function parameter:

let p = 1;

function myFunc(q = p + 1) {
  return q;
}

myFunc(); // => 2

The function parameter is renamed from

p

to q. When myFunc() is called, no parameters are specified, so the parameter q is initialized to the default value p 1. To evaluate p 1, access the variable p of the outer scope: p 1 = 1 1 = 2. <h3 id="strong-函数声明与类声明-strong"><strong>5. 函数声明与类声明</strong></h3> <p>以下代码在代码块内定义了一个函数和一个类：</p><pre class='brush:php;toolbar:false;'>if (true) { function greet() { // function body } class Greeter { // class body } } greet(); // ??? new Greeter(); // ???</pre><p>是否可以在块作用域之外访问 <code>greet 和 Greeter？ （考虑 ES2015 环境）

答案

function 和 class 声明都是块作用域的。所以在代码块作用域外调用函数 greet() 和构造函数 new Greeter() 就会抛出 ReferenceError。

6. 总结

必须注意 var 变量，因为它们是函数作用域的，即使是在代码块中定义的。

由于 ES2015 模块系统是静态的，因此你必须在模块作用域内使用 import 语法（以及 export）。

函数参数具有其作用域。设置默认参数值时，请确保默认表达式内的变量已经用值初始化。

在 ES2015 运行时环境中，函数和类声明是块作用域的。但是在 ES2015 之前的环境中，函数声明仅在函数作用域内。

希望这些陷阱能够帮你巩固作用域知识！

英文原文地址：https://dmitripavlutin.com/javascript-scope-gotchas/

作者：Dmitri Pavlutin

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