为什么函数结束后我的数组变得未定义？

debugcn 发表于 Dev

FireyDeath4-YuAre一个陌生人

好的，我只是在做一个将所有内容从一个列表移到另一个列表的函数，完成后，它会自发地变得不确定。这样做没有充分的理由，但是不好的原因是什么呢？

来看

function moveList(){//in my real program it actually shuffles the list, this one just transfers the contents
    var listOne=[0, 1, 2, 3];//we have a list
    var listTwo=[];//now we have an empty list
    var length = listOne.length//and now we have the list length
    for (run = 0; run < length; run++) {
        listTwo.push(listOne[0]);//it just copies the first entry from the first list to the second
        listOne.splice(0, 1,)//and deletes it from the first list
    }//this does it for an entry in the list, it's length amount of times
    console.log("listTwo: " + listTwo);//this tells us what the new list now is, and it works
    console.log("listOne: " + listOne);//this tells us what the original list is, which is empty
}
var listOne;
var listTwo;
//these are mandatory, without this, even with the function below script,
//you have Uncaught ReferenceError: list(One and Two) is not defined
moveList();//now we do the function, and then,
console.log("listTwo: " + listTwo);//THE LIST IS NOW UNDEFINED??? What??????

只是为什么我只是在我的函数中定义了它。你怎么了（从字面上看，我没有看到）？

艾哈迈德·哈默德|

原因是您用listTwo不同范围的名称初始化了两个列表！全局作用域中的一个变量，您不给它赋值，因此它停留在undefined，而另一个变量在函数内部，并且不能从函数外部访问，因为已经有另一个具有相同名称的变量！

为了使代码正常工作，您应该在函数内部使用变量而不声明它们。

function moveList(){//in my real program it actually shuffles the list, this one just transfers the contents
    listOne=[0, 1, 2, 3];//we have a list
    listTwo=[];//now we have an empty list
    var length = listOne.length//and now we have the list length
    for (run = 0; run < length; run++) {
        listTwo.push(listOne[0]);//it just copies the first entry from the first list to the second
        listOne.splice(0, 1,)//and deletes it from the first list
    }//this does it for an entry in the list, it's length amount of times
    console.log("listTwo: " + listTwo);//this tells us what the new list now is, and it works
    console.log("listOne: " + listOne);//this tells us what the original list is, which is empty
}
var listOne;
var listTwo;

moveList();//now we do the function, and then,
console.log("listTwo: " + listTwo);//THE LIST IS NOW NOT UNDEFINED ;)

同样适用于listOne变量。

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