当var出现之后，为什么函数变量未定义？

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问题描述

我已经读过一个好的做法是放置一个var语句，它定义了每个函数顶部的所有局部变量。下面的代码说明了为什么这是一个好主意，因为显然使用变量后 em 未定义。

I've read that a good practice is to put a var statement which defines all local variables at the top of each function. The following code shows why this is a good idea, since apparently a var after a variable is used makes it undefined.

但是有人可以告诉我为什么这种情况？

But can someone tell me why this is the case?

<html>
    <head>
        <script type="text/javascript">
            window.onload = function() {
                var a=1;
                function foo() {
                    //a = 2; //outputs 2,2 because this overwrites the external variable value
                    //var a = 2; //outputs 2,1 because the "var" keyword creates a second variable with local scope which doesn't affect the external variable
                    console.log(a);
                    var a = 3; //ouputs "undefined,1" ???
                }
                foo();
                console.log(a);
            };
        </script>
    </head>
    <body>

    </body>
</html>

推荐答案

function foo() {
  console.log(a);
  var a = 3;
}

相当于

function foo() {
  var a;
  console.log(a);
  a = 3;
}

因为在JavaScript变量声明被悬挂但是初始值设定项不是。

because in JavaScript variable declarations are hoisted but initializers are not.

您可以通过以下示例看到这是真的：

You can see that this is literally true with the following example:

e = 0;
function foo() {
  e = 1;
  try {
    throw 2;
  } catch (e) {
    var e = 3;
    alert("In catch " + e);
  }
  alert("Before end of function " + e);
}
foo();
alert("Outside function " + e);

提醒

因为变量声明已被提升，因此函数外部的 e 不会被 e = 1 更改，但 e = 3 发生在 catch 内，所以 3 不会影响函数末尾的 e ，而不是覆盖异常值。

because the variable declaration is hoisted so the e outside the function is not changed by e = 1, but the e = 3 occurs inside the catch so the 3 does not affect the e at the end of the function, instead over-writing the exception value.

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