prototype method is not a function,why?

Question

(function(){
 test();
}());
function Class(){
 this.prop = 'hi';
}
Class.prototype.mod = function(num){this.prop = num;}
function test(){
 var c = new Class();
 c.mod('now'); // it'll say it's not a function
 alert(c.prop); // it's work
}

I wanna move function and class out to ready function to make code clean up and save memory, but I found the class method does not work.

If I moved prototype to test function, it work, like

(function(){
 test();
}());
function Class(){
 this.prop = 'hi';
}
function test(){
 Class.prototype.mod = function(num){this.prop = num;}
 var c = new Class();
 c.mod('now'); // it's ok
 alert(c.prop); 
}

why I must to move prototype method to test or ready function?

• JJJ: The fact that you call it "ready function" seems to suggest you think the self-executing function ((function(){...}())) waits for the page to load before executing, which is not true.
• Roy: as you say, it's self-executing function and I make this js script bottom of page, like ready. is it good to make everything in self-executing function I wanna know, thx

Answer 1

Because your .prototype.mod definition is after the function that calls it. Hoisting only helps for the function definition itself (which is why new Class() works fine), not for prototype definitions.

This really shouldn't be so hard: prepare your tools first, then use them.

• Roy: you're right, btw do you have any way to hoist .prototype?