如何在变量的闭包内调用函数？

Question

I'm a newbie in java but the code is below the explanation This is code from a game but the problem is: Take the navigator function, I click the navigator in the game and some sort of (ajax?) goes on and it gets logged. But when I use my FireBug console I cannot send a message with the functions above the return line. Essentially.. The only functions that ThisFunction.* shows is the ones returned at the bottom. How can I invoke the Navigator function? I've tried: ThisFunction.a.navigator(args here);

, but it says a is undefined.. it doesn't show in the autocomplete list either.

** I removed the code because it is from a game. Thanks for the help! **

Answer 1

You somewhat nailed it on its head with this bit:

The only functions that ThisFunction.* shows is the ones returned at the bottom

That is the expected and purposeful functioning of the language.

Short answer: You have to return out of the closure anything that you want to be externally accessible... That could be the a variable, or it could be an api that could itself access the a variable while keeping it private from the exterior. This is called lexical scoping and it is your friend.

Example time:

var ThisFunction = (function() {
  var a = { navigator: "woot" };
  var b = function() {
    return a;
  }
});
ThisFunction.a; //a is null/undefined on the returned
ThisFunction.b; //b is defined yay
var aOUTSIDE = ThisFunction.b();
aOUTSIDE.navigator; // "woot"

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Closures

So the only things that can get at A are the things that were var'd up in the same "scope" as a. So either you return A out of the scope or you return something from inside the scope that provides an API to either get at A, or to execute some of A's internals...

Answer 2

Depending on what OTHERFUNCTIONSHERE is, you can access it from within one of those functions, if they close over the variable a (why such cryptic var names by the way?). Otherwise, it's out of scope.

Answer 3

Given that you're using Firebug, ThisFunction.%a.navigator(... args) should work ( .% is a Firebug-specific extension to the language). But as noted in other answers, it's impossible from pure JavaScript.