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The formula 🤨 : four bases, two empty objects, and five falsy values.

Seven data types

  • number string bool symbol
  • null undefined
  • object

Five falsy value

  • null undefined
  • 0 NaN
  • “(empty string)

preface

When executing a JS function, the result will be different depending on the position of the function.

Let’s look at a few examples of how the timing of a function’s execution can affect the result.

The body of the

Case 1

let a = 1
function fn(){
    console.log(a)
}
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Tang’s monk: apprentice son, excuse me on regular meeting print out how many?

Wukong: I don’t know, because you didn’t call function fn.

Case 2

let a = 1
function fn(){
    console.log(a)
}
fn()
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Tang’s monk: apprentice son, excuse me on regular meeting print out how many?

Obviously, it is 1.

Example 3

let a = 1
function fn(){
    console.log(a)
}
a = 2
fn()
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Tang’s monk: apprentice son, excuse me on regular meeting print out how many?

Shaseng: Because fn is executed after a=2, print 2.

Example 4

let a = 1
function fn(){
    console.log(a)
}
fn()
a = 2
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Tang’s monk: apprentice son, excuse me on regular meeting print out how many?

White dragon: Comparing with example 3, we can see that fn is called first, so it will print 1.

The first few examples are relatively simple, I believe you can answer. So now let’s upgrade the difficulty, look at this:

Is the

Different timing, different results

   let i = 0
   for(i = 0; i<6; i++){
     setTimeout(() = >{
       console.log(i)
     },0)}// Output six 6's
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The for loop is in the main thread, and setTimeout is an asynchronous method. In the task queue, the task queue will execute only after the main thread has finished executing. When the for is finished, the I value is already 6, so the result is 6. So how do I print 0 minus 5.

   for(let i = 0; i<6; i++){
     setTimeout(() = >{
       console.log(i)
     },0)}Copy the code

This is one of the things that JS changed later for new developers. Let the code do what the new guy wants. So what else can we do to print 0 to 5?

This can be done by self-executing functions

     for (var i = 0; i <= 5; i++) { ! (function (j) {
          setTimeout(function print() {
            console.log(j);
          }, 0);
        })(i);
      }
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  1. By setTimeout the third argument
    for (var i = 0; i < 6; i++) {
        setTimeout((i) = > console.log(i), 0, i);
      }
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It took me two months to understand let

As my level is limited, if there is an inaccurate description of the place please give me a message, welcome to exchange ~

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