javascript - javascript 闭包的困惑

Question

我已被证明我并不真正理解javascript闭包，并且我对以下代码感到困惑。我以为 fxn 会访问外部 foo，但它实际上打印出“underfined”。为什么？？

var foo = "hello";
function fxn(){
   alert(foo);
   var foo = "test"
}

fxn();

Answer 1

这是因为在 JavaScript 中，变量得到 hoisted ，这意味着

Variables are initialised to undefined when created. A variable with an Initialiser is assigned the value of its AssignmentExpression when the VariableStatement is executed, not when the variable is created._{(ES5 §12.2)}

因此，从语义上讲，您的代码将等同于以下内容......

var foo = "hello";
function fxn(){
   var foo; //Variables are initialised to undefined when created
   alert(foo);
   foo = "test"; //A variable with an *Initialiser* is assigned the value of its *AssignmentExpression* when the *VariableStatement* is **executed**
}

fxn();