A summary of tips on how to avoid common pitfalls in JavaScript

1. Have you tried sorting the array elements?

JavaScript uses alphanumeric order by default. Therefore, the result of [1,2,5,10].sort() is [1, 10, 2, 5].

If you want to sort correctly, you should do this: [1,2,5,10].sort((a, b) => a - b)

2. new Date() is very easy to use

new Date() can receive:

• - does not receive any parameters: returns the current time;

• - Receives a parameter `x`: Returns the value of January 1, 1970 + `x` milliseconds.

• - `new Date(1, 1, 1)` returns February 1, 1901.

• - However...., `new Date(2016, 1, 1)` does not add 2016 to 1900, but only represents 2016.
  

#3. The replacement function does not really replace?

let s = "bob"
const replaced = s.replace('b', 'l')
replaced === "lob" // 只会替换掉第一个b
s === "bob" // 并且s的值不会变

If you want to replace all b, use regular expressions:

"bob".replace(/b/g, 'l') === 'lol'

4. Be careful with comparison operations

// 这些可以
'abc' === 'abc' // true
1 === 1 // true
// 然而这些不行
[1,2,3] === [1,2,3] // false
{a: 1} === {a: 1} // false
{} === {} // false

Because [1,2,3] and [1,2,3] are two different arrays, but their elements happen to be the same. Therefore, it cannot be judged simply by `===`.

5. Array is not a basic type

typeof {} === 'object' // true
typeof 'a' === 'string' // true
typeof 1 === number // true
// 但是....
typeof [] === 'object' // true

If you want to determine whether a variable `var` is an array, you need to use `Array.isArray(var)` .

6. Closure

This is a classic JavaScript interview question:

const Greeters = []
for (var i = 0 ; i < 10 ; i++) {
 Greeters.push(function () { return console.log(i) })
}
Greeters[0]() // 10
Greeters[1]() // 10
Greeters[2]() // 10

Although the expected output is 0,1,2,..., the actual output is But it won't. Do you know how to debug?

There are two ways:

• - Use `let` instead of `var`.

• - Use the `bind` function.

Greeters.push(console.log.bind(null, i))

Of course, there are many solutions. These two are my favorite!

7. Regarding bind

What will the following code output?

class Foo {
 constructor (name) {
 this.name = name
 }
 greet () {
 console.log('hello, this is ', this.name)
 }
 someThingAsync () {
 return Promise.resolve()
 }
 asyncGreet () {
 this.someThingAsync()
 .then(this.greet)
 }
}
new Foo('dog').asyncGreet()

If you say that the program will crash and report an error: Cannot read property 'name' of undefined.

Because the `geet` on line 16 is not executed in the correct environment. Of course, there are many ways to solve this BUG!

- Prefer to use the `bind` function to solve the problem:

asyncGreet () {
 this.someThingAsync()
 .then(this.greet.bind(this))
}

This will ensure that `greet` will be called by the instance of Foo, rather than the local function `this`.

- If you want `greet` to never be bound to the wrong scope, you can use `bind` inside the constructor.

class Foo {
 constructor (name) {
 this.name = name
 this.greet = this.greet.bind(this)
 }
}

- You can also use arrow functions (=>) to prevent the scope from being modified.

asyncGreet () {
 this.someThingAsync()
 .then(() => {
 this.greet()
 })
}

8. Math.min() is larger than Math.max()

Math.min() < Math.max() // false

Because Math.min() returns Infinity, while Math.max ()Return -Infinity.

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