The first thing you need to understand is that the task mechanism of JS is the queue mechanism.

So instead of running a for loop, we just put the setTimeout task at the back of the queue. That is, the code executed in setTimeout will only be executed after the for loop is executed, so the value of i at that time is the value that does not satisfy the for loop. , will execute the setTimeout code.

Personal opinion, please point out if there is anything wrong

var can be changed to let

var is a global definition, i does not form a closure, and log(i) prints the final value of i 6

let is a block-level domain

for (let i = 1; i <= 5; i++) { setTimeout(function timer() { console.log(i); }, i * 1000); }

setTimeout has two characteristics, its this is separated from the context's this, and its call is asynchronous.

This is caused by [asynchronous], the for loop will be completed first, and then setTimeout will be executed, because the for loop will be 6 at the end every time it is executed - so naturally when setTimeout calls i, it will be five 6

Solution 1:

var 换 let

Solution 2:

楼主自己用的闭包

Solution 3:

//类似于方法二，使用setTimeout的第三个参数直接传参 for (var i = 1; i <= 5; i++) { setTimeout(function(i) { console.log(i); }, i * 1000,i); }