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。

这个解释是从我在Medium上写的一篇文章中得出的:

Hoisting 是一个 JavaScript 机制,在其中变量和函数声明被转移到其范围的顶部,由分数者将源代码列入一个中间代表性之前,实际的代码执行由 JavaScript 解释器开始。

我想将这些关键词链接到执行背景,因为执行背景在这一切中都很重要。执行背景有两个阶段:创作阶段和执行阶段。

在创建阶段的执行背景, var, let 和 const 将仍然存储其变量在记忆中与未确定的值在该执行背景的变量环境中. 差异在执行阶段. 如果您使用参考一个变量定义为 var 之前它被分配一个值,它将只是不确定的。

function a(){
    b;
    let b;
}
a();
> Uncaught ReferenceError: b is not defined

使用 var 声明定义的变量在其定义的整个函数中已知,从函数的开始。 (*) 使用 let 声明定义的变量仅在其定义的区块中已知,从其定义的时刻开始。

// i IS NOT known here
// j IS NOT known here
// k IS known here, but undefined
// l IS NOT known here

function loop(arr) {
    // i IS known here, but undefined
    // j IS NOT known here
    // k IS known here, but has a value only the second time loop is called
    // l IS NOT known here

    for( var i = 0; i < arr.length; i++ ) {
        // i IS known here, and has a value
        // j IS NOT known here
        // k IS known here, but has a value only the second time loop is called
        // l IS NOT known here
    };

    // i IS known here, and has a value
    // j IS NOT known here
    // k IS known here, but has a value only the second time loop is called
    // l IS NOT known here

    for( let j = 0; j < arr.length; j++ ) {
        // i IS known here, and has a value
        // j IS known here, and has a value
        // k IS known here, but has a value only the second time loop is called
        // l IS NOT known here
    };

    // i IS known here, and has a value
    // j IS NOT known here
    // k IS known here, but has a value only the second time loop is called
    // l IS NOT known here
}

loop([1,2,3,4]);

for( var k = 0; k < arr.length; k++ ) {
    // i IS NOT known here
    // j IS NOT known here
    // k IS known here, and has a value
    // l IS NOT known here
};

for( let l = 0; l < arr.length; l++ ) {
    // i IS NOT known here
    // j IS NOT known here
    // k IS known here, and has a value
    // l IS known here, and has a value
};

loop([1,2,3,4]);

// i IS NOT known here
// j IS NOT known here
// k IS known here, and has a value
// l IS NOT known here

今天使用安全吗?

有些人会说,在未来,我们只会使用让陈述,而这些陈述会变得过时。JavaScript老师Kyle Simpson写了一篇非常复杂的文章,他认为为什么不会这样。

事实上,我们实际上需要问自己是否安全使用放弃声明,这个问题的答案取决于你的环境:

此分類上一篇

如何跟踪浏览器支持

// An array of adder functions.
var adderFunctions = [];

for (var i = 0; i < 1000; i++) {
  // We want the function at index i to add the index to its argument.
  adderFunctions[i] = function(x) {
    // What is i bound to here?
    return x + i;
  };
}

var add12 = adderFunctions[12];

// Uh oh. The function is bound to i in the outer scope, which is currently 1000.
console.log(add12(8) === 20); // => false
console.log(add12(8) === 1008); // => true
console.log(i); // => 1000

// It gets worse.
i = -8;
console.log(add12(8) === 0); // => true

上面的过程不会产生所需的函数序列,因为我的范围超越了每个函数创建的区块的 iteration。 相反,在环节结束时,每个函数的 i 关闭指在环节结束时的 i 值(1000)为每个在 adder 中的匿名函数。

// Let's try this again.
// NOTE: We're using another ES6 keyword, const, for values that won't
// be reassigned. const and let have similar scoping behavior.
const adderFunctions = [];

for (let i = 0; i < 1000; i++) {
  // NOTE: We're using the newer arrow function syntax this time, but 
  // using the "function(x) { ..." syntax from the previous example 
  // here would not change the behavior shown.
  adderFunctions[i] = x => x + i;
}

const add12 = adderFunctions[12];

// Yay! The behavior is as expected. 
console.log(add12(8) === 20); // => true

// i's scope doesn't extend outside the for loop.
console.log(i); // => ReferenceError: i is not defined

每个函数现在保留在函数创建时的 i 的值,并且 adderFunctions 按照预期行事。

现在,图像将两种行为混合在一起,你可能会看到为什么不建议在同一脚本中混合更新的Let和 const。

const doubleAdderFunctions = [];

for (var i = 0; i < 1000; i++) {
    const j = i;
    doubleAdderFunctions[i] = x => x + i + j;
}

const add18 = doubleAdderFunctions[9];
const add24 = doubleAdderFunctions[12];

// It's not fun debugging situations like this, especially when the
// code is more complex than in this example.
console.log(add18(24) === 42); // => false
console.log(add24(18) === 42); // => false
console.log(add18(24) === add24(18)); // => false
console.log(add18(24) === 2018); // => false
console.log(add24(18) === 2018); // => false
console.log(add18(24) === 1033); // => true
console.log(add24(18) === 1030); // => true

不要让这件事发生在你身上,使用灯具。

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