为什么这种关闭工作？

Question

Say I have a simple function that alerts a message:

function callMessage(msg){
        alert(msg);
    }

Now when I call it like so, it does not work. Throws error "hey is not defined"

function sayHi(){
        var hey = "hi there"
        setTimeout("callMessage(hey)", 1000);
    }
    sayHi();

But when I call it inside an anonymous function it does work:

function sayHi(){
        var hey = "hi there"
        setTimeout(function(){callMessage(hey);}, 1000);
    }
    sayHi();

Why is the "hey" variable only visible when I put it inside an anonymous function?

Answer 1

In the first example, the code is evaluated after the timer expired and the current scope was left. hey is undefined at that moment.

The second example - the proper way to use setTimeout - uses an anonymous function created when invoking setTimeout() . This anonymous function also receives a copy of the current scope.

Answer 2

"callMessage(hey)" is a string, not a closure. It's evaluated when the timeout runs, and at this point the variable hey isn't in scope.

Answer 3

it's normal.

The second example creates what we call a fixture, this is an execution context. hey variable variable is saved to be used by the anonymous function in memory.

In your first example, the hey variable is not saved in a fixture (because javascript can't know that you will used the variable after) and so can not be retrieved when the string is evaluated