javascript 覆盖javascript函数中的变量
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override variable in javascript function
提问by elevenMinutes
Firstly I want to say that I was searching some information about scope variables in javascript (e.g What is the scope of variables in JavaScript?) and I was trying to understand why my code behaves like that but finally I really don't know. I can't override var a in my function in function, I can do this only using a = 6;without var. Why a is undefined before if statement and overriding this variable ? Why there is no result like this:
首先，我想说我在 javascript 中搜索有关范围变量的一些信息（例如JavaScript 中变量的范围是什么？），我试图理解为什么我的代码会这样，但最后我真的不知道。我不能在我的函数中覆盖 var a，我只能在a = 6;没有 var 的情况下使用它。为什么 a 在 if 语句之前未定义并覆盖此变量？为什么没有这样的结果：
5
equal 5
equal 6
5
等于 5
等于 6
I have something like this instead:
我有这样的事情：
undefined
not equal 5
equal 6
不明确的
不等于 5
等于 6
Here's my code:
这是我的代码：
var a = 5;
function something(){
console.log(a);
if(a == 5){
console.log('equal 5')
}
else{
console.log('not equal 5');
}
var a = 6;
function test(){
if(a == 6){
console.log('equal 6');
}
else{
console.log('not equal 6')
}
}
test();
}
something();
回答by xdazz
This is because javascript variable hoisting.
这是因为 javascript 变量提升。
Your code is equal to:
您的代码等于：
var a = 5;
function something(){
var a; // here
console.log(a); // a is the local variable which is undefined
if(a == 5){
console.log('equal 5')
}
else{
console.log('not equal 5'); // a is undefined so not equal 5
}
a = 6; // a is assigned to 6
function test(){
if(a == 6){
console.log('equal 6'); // a is 6 here
}
else{
console.log('not equal 6')
}
}
test();
}
something();
回答by cerpintaxt
Because when you declare var a = 6the second time you are overriding the first declaration of a, i.e. var a = 5. So when you define the function something(), ahas not yet been defined, hence console.log(a)returns undefined.
因为当您var a = 6第二次声明时，您将覆盖a，即的第一次声明var a = 5。因此，当您定义函数时something()，a尚未定义，因此console.log(a)返回未定义。
However when you replace the second declaration of awith just a = 6the first declaration remains valid, you just change the value of aafterwards and so you get the expected result:
但是，当您a仅用a = 6第一个声明替换第二个声明时仍然有效，您只需更改 after 的值即可a获得预期结果：
5
equal 5
equal 6
See this postfor more discussion.
有关更多讨论，请参阅此帖子。
