// 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。

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

// 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。

下面表明“让我们”和“存在”在范围内是如何不同的:

let gfoo = 123;
if (true) {
    let gfoo = 456;
}
console.log(gfoo); // 123

var hfoo = 123;
if (true) {
    var hfoo = 456;
}
console.log(hfoo); // 456

gfoo,定义为 let 最初是在全球范围内,当我们宣布 gfoo 再次在如果条款的范围改变,当一个新的值被分配到变量在该范围内,它不会影响全球范围。

虽然 hfoo,由 var 定义,起初在全球范围内,但再次当我们在 if 条款内宣布它时,它考虑到全球范围 hfoo,尽管 var 已再次被用来宣布它。

函数 VS 区块范围:

与 var 和 let 的主要区别是,与 var 宣言的变量是函数分解的,而与 let 宣言的函数是区块分解的。

function testVar () {
  if(true) {
    var foo = 'foo';
  }

  console.log(foo);
}

testVar();  
// logs 'foo'


function testLet () {
  if(true) {
    let bar = 'bar';
  }

  console.log(bar);
}

testLet(); 
// reference error
// bar is scoped to the block of the if statement 

当第一个函数测试Var 被称为变量 foo 时,与 var 声明,仍然在 if 声明之外可用。

与Let的变量:

当第二个函数测试Let被称为变量栏,用Let表示,仅可在 if 声明中访问,因为用Let表示的变量是区块分开的(其中一个区块是曲线条之间的代码,例如,如果{},为{},函数{})。

请不要让变量变量:

变量与不要被抓住:

console.log(letVar);

let letVar = 10;
// referenceError, the variable doesn't get hoisted

与 var do 相容的变量:

console.log(varVar);

var varVar = 10;
// logs undefined, the variable gets hoisted

var bar = 5;
let foo  = 10;

console.log(bar); // logs 5
console.log(foo); // logs 10

console.log(window.bar);  
// logs 5, variable added to window object

console.log(window.foo);
// logs undefined, variable not added to window object

让它很有趣,因为它允许我们做这样的事情:

(() => {
    var count = 0;

    for (let i = 0; i < 2; ++i) {
        for (let i = 0; i < 2; ++i) {
            for (let i = 0; i < 2; ++i) {
                console.log(count++);
            }
        }
    }
})();

其次,有数以数以数。

在哪里

(() => {
    var count = 0;

    for (var i = 0; i < 2; ++i) {
        for (var i = 0; i < 2; ++i) {
            for (var i = 0; i < 2; ++i) {
                console.log(count++);
            }
        }
    }
})();

只有数数(0,1 )。