为什么函数结束后我的数组就变得未定义?

Question

好吧，我只是在做一个将所有内容从一个列表移动到另一个列表的函数，一旦完成，它就会自发地变得未定义。它没有充分的理由这样做，但坏理由是什么？

在这里，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??????

为什么呢？我真的只是在我的函数中定义了它。你怎么了(字面意思，我看不出来(？

Answer 1

原因是您初始化了两个名称为listTwo在不同范围内的列表！一个在全局范围内，你不给它一个值，所以它保持undefined，另一个在函数内，并且无法从函数外部访问，因为已经有另一个同名的变量！

为了使您的代码正常工作，您应该使用函数中的变量而不声明它们。

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 ;)

这同样适用于listOne变量。

Answer 2

您有两个单独的listTwo变量。一个是全局的，而另一个是函数moveList的局部。这两种事件彼此无关。listOne也是如此.

所以你的代码等效于下面的代码，我把名字改得更清楚：

function moveList(){//in my real program it actually shuffles the list, this one just transfers the contents
var localListOne=[0, 1, 2, 3];//we have a list
var localListTwo=[];//now we have an empty list
var length = localListOne.length//and now we have the list length
for (run = 0; run < length; run++) {
localListTwo.push(localListOne[0]);//it just copies the first entry from the first list to the second
localListOne.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: " + localListTwo);//this tells us what the new list now is, and it works
console.log("listOne: " + localListOne);//this tells us what the original list is, which is empty
}
var globalListOne;
var globalListTwo;
//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: " + globalListTwo); // Obviously globalListTwo has never been given a value, so it is undefined.