索引为"for".在"setInterval"内部 [英] Index of "for" inside "setInterval"

问题描述

我有一个对象，并将其放入一个数组中，并使用 for 迭代该数组，对于数组中的每个对象，我都设置了一个设置间隔，以使用一个参数，但我不能.这是我的班级".

I have an object and i put it in a Array, and the i iterate the array with for and for each object in the array i put a set interval for calling a method of the object with a parameter, but i cant. This is my "Class".

    function ClassTest() {
        this.test = function(word) {
            console.log(word);
        }
    }

我创建对象并将其放在数组上:

The i create the object and put it on a Array:

    var array = [];
    var objectTest = new ClassTest();


    array.push(objectTest);

当我设置时间间隔时:

    for(var i = 0; i < array.length; i++) {
        array[i].loop = setInterval(function() {
            array[i].test("hello")
        }, 1000);
    }

问题是setInterval函数主体中的var i 存在，我可以创建一个var index 并且存在，但是我不明白为什么 index存在，而var i 不存在:

The problem is that the var i in the setInterval function dosent exist, i can create a var index and the it exist but i dont understand why index exist and the var i not.:

    for(var i = 0; i < array.length; i++) {
        var index = i;
        array[i].loop = setInterval(function() {
            array[index].test("hello")
        }, 1000);
    }

当我不使用 index var时，会出现此错误:

I get this error when i dont use the index var:

Uncaught TypeError: Cannot read property 'test' of undefined

推荐答案

因为您的 for 循环会立即执行.到您的 setInterval()首次执行时，您的 for 循环将持续很长时间，并且您的 i 将不在数组范围.

Because your for loop executes immediately. By the time your setInterval() executes for the first time, your for loop will have long since finished and your i will be outside of the range of your array.

在此示例中，我们的 array 的 length 为3.当我们的 for 循环结束时，我们的 i 变量将等于 3 .如果我们在 setTimeout :

In this example, our array has a length of 3. When our for loop finishes, our i variable will be equal to 3. The number 3 will be displayed three times in our JavaScript console if we log i within the setTimeout:

var array = [1, 2, 3];

for (var i = 0; i < array.length; i++) {
  setTimeout(function() {
    console.log(i);
  }, 1000);
}

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