让

区块范围

在顶级(功能之外)

var globalVariable = 42;
let blockScopedVariable = 43;

console.log(globalVariable); // 42
console.log(blockScopedVariable); // 43

console.log(this.globalVariable); // 42
console.log(this.blockScopedVariable); // undefined

在一个功能之内

(() => {
  var functionScopedVariable = 42;
  let blockScopedVariable = 43;

  console.log(functionScopedVariable); // 42
  console.log(blockScopedVariable); // 43
})();

console.log(functionScopedVariable); // ReferenceError: functionScopedVariable is not defined
console.log(blockScopedVariable); // ReferenceError: blockScopedVariable is not defined

在一个区块内

{
  var globalVariable = 42;
  let blockScopedVariable = 43;
  console.log(globalVariable); // 42
  console.log(blockScopedVariable); // 43
}

console.log(globalVariable); // 42
console.log(blockScopedVariable); // ReferenceError: blockScopedVariable is not defined

在一个圈子里

用Let in Loops宣言的变量只能在该圈内提到。

for (var i = 0; i < 3; i++) {
  var j = i * 2;
}
console.log(i); // 3
console.log(j); // 4

for (let k = 0; k < 3; k++) {
  let l = k * 2;
}
console.log(typeof k); // undefined
console.log(typeof l); // undefined
// Trying to do console.log(k) or console.log(l) here would throw a ReferenceError.

// Logs 3 thrice, not what we meant.
for (var i = 0; i < 3; i++) {
  setTimeout(() => console.log(i), 0);
}

// Logs 0, 1 and 2, as expected.
for (let j = 0; j < 3; j++) {
  setTimeout(() => console.log(j), 0);
}

临时死区

由于临时死亡区,使用声明的变量在被声明之前无法访问,试图这样做会导致错误。

console.log(noTDZ); // undefined
var noTDZ = 43;
console.log(hasTDZ); // ReferenceError: hasTDZ is not defined
let hasTDZ = 42;

没有重新宣布

var a;
var a; // Works fine.

let b;
let b; // SyntaxError: Identifier 'b' has already been declared

var c;
let c; // SyntaxError: Identifier 'c' has already been declared

格斯特

没有重新分配

const a = 42;
a = 43; // TypeError: Assignment to constant variable.

请注意,这并不意味着值是不可改变的;它的属性仍然可以改变。

const obj = {};
obj.a = 42;
console.log(obj.a); // 42

如果您想要一个不可变的对象,您应该使用 Object.freeze()。

const obj = Object.freeze({a: 40});
obj.a = 42;
console.log(obj.a); // 40
console.log(obj.b); // undefined

初步者需要

在使用 const 表示变量时,您必须始终指定值。

const a; // SyntaxError: Missing initializer in const declaration

在 MDN 中查看此链接

let x = 1;

if (x === 1) {
let x = 2;

console.log(x);
// expected output: 2
}

console.log(x);
// expected output: 1

函数运行() { var foo = “Foo”; let bar = “Bar”; console.log(foo, bar); // Foo Bar { var Moo = “Mooo” let baz = “Bazz”; console.log(moo, baz); // Mooo Bazz } console.log(moo); // Mooo console.log(baz); // ReferenceError } run();

為什麼讓關鍵字被引入到語言是功能範圍是混亂的原因,是JavaScript的主要錯誤來源之一。

创建全球对象财产

在最高层次上,让我们不同于VAR,不会在全球对象上创造任何财产:

var foo = “Foo”; // 全球推翻的Let bar = “Bar”; // 不允许全球推翻的console.log(window.foo); // Foo console.log(window.bar); // undefined

const name = 'Max'; let age = 33; var hasHobbies = true; name = 'Maximilian'; age = 34; hasHobbies = false; const summarizeUser = (userName, userAge, userHasHobby) => { return ( 'Name is'+ userName + ', age is'+ userAge +'and the user has hobbies:'+ userHasHobby ); } console.log(summarizeUser(name, age, hasHobbies));

正如你可以从上面的代码运行中看到的那样,当你尝试更改 const 变量时,你会发现一个错误:

试图超越一个“名称”,这是一个恒定的。

或

TypeError: Invalid assignment to const 'name'.

但是,看看放变量。

首先,我们宣布让年龄=33岁,然后将另一个值年龄=34岁,这是OK;当我们试图改变时,我们没有任何错误。

msg = “Hello World” 函数 doWork() { // msg 将是可用的,因为它被定义在这个开幕式上! 让朋友 = 0; console.log(msg); // 与 VAR 虽然: for (var iCount2 = 0; iCount2 < 5; iCount2++) {} // iCount2 将在这个开幕式后可用! console.log(iCount2); for (let iCount1 = 0; iCount1 < 5; iCount1++) {} // iCount1 将没有

答答答答答答答答答答答答答答答答答

var 变量是全球性的,基本上可以到达到任何地方,而让变量不是全球性的,只有直到一个关闭的偏见杀死它们。

请参见下面的我的例子,并注意狮子(Let)变量如何在两个 console.logs 中以不同的方式行事;它在第二个 console.log 中变得无效。

var cat = "cat";
let dog = "dog";

var animals = () => {
    var giraffe = "giraffe";
    let lion = "lion";

    console.log(cat);  //will print 'cat'.
    console.log(dog);  //will print 'dog', because dog was declared outside this function (like var cat).

    console.log(giraffe); //will print 'giraffe'.
    console.log(lion); //will print 'lion', as lion is within scope.
}

console.log(giraffe); //will print 'giraffe', as giraffe is a global variable (var).
console.log(lion); //will print UNDEFINED, as lion is a 'let' variable and is now out of scope.