函数中未定义的变量

Question

Look at the following snippet:

var fruit = "Orange";
echoThis();

function echoThis() {    
    alert(fruit);
    var fruit = "Apple";
}

When I run this snippet, the alert of fruit is undefined . Why?

First I thought it has something to do with hoisting, that within a funciton, the JS engine "lifts" all var declarations to the top or something, but then I would expect the alert to display Apple , not undefined .

There's probably some elementary JS behaviour I'm unaware of. Anyone care to explain?

JS fiddle: https://jsfiddle.net/z37230c9/

Answer 1

It's because of hoisting . Variables declared in a function are available for the entire scope, but they are assigned a value where you put it.

This:

function echoThis() {    
    alert(fruit);
    var fruit = "Apple";
}

becomes this:

function echoThis() {    
   var fruit;
   alert(fruit);
   fruit = "Apple";
}

This is why you code is evaluated successfully, but fruit's value is undefined.

also:

var fruit = "Orange"; //this fruit variable
echoThis();

function echoThis() {    
    alert(fruit);
    var fruit = "Apple"; // is overridden by this, 
     //because it's re-declared in a local scope. 
}

if you really want to change this then remove the var in the function.

   function echoThis() {    
        alert(fruit);
        fruit = "Apple";  //this now accesses the global fruit.
    }

Answer 2

The variable is hoisted to the top of the function when it is compiled.

function echoThis() {    
    alert(fruit);
    var fruit = "Apple";
}

Translates to this

function echoThis() {    
    var fruit;
    alert(fruit);
    fruit = "Apple";
}

So when the alert is called it is technically undefined at this point. To stop seeing that move your variable declaration before the alert statement.

Answer 3

that's because js has lexical scope and hoisting.

so echoThis looks for fruit within itself before looking outside and since you have var fruit = "Apple"; , fruit is hoisted in echoThis .

Answer 4

Besides the above responses, if you want to get the orange AND the apple after.. you could try to pass the variable simply as a parameter and work in the function

 var fruit = "Orange"; echoThis(fruit); function echoThis(fruit) { alert(fruit); // You will get the orange var fruit = "Apple"; alert(fruit); // You will get the apple }