Why can't I declare local variables and functions within a TypeScript class?
javascript, typescript
Solution
As apsillers mentions, `private static` is probably what you want. While it's not supported in current builds, you will be able to have a `private static` member in TypeScript at some point in the future (the design team changed its mind on this one based on feedback similar to this).
Problem
In TypeScript, I can't seem to declare a function in a class without the compiler adding it to the prototype. For example: ``` class MyTypeScriptClass { // method, is added to prototype foo1(): void { alert('invoked foo1'); } // private method also added to prototype private foo2(): void { alert('invoked foo2'); } //// can I have a local function, without making it a private method? //function foo3() { // alert('invoked foo3'); //} } ``` The above compiles to this: ``` var MyTypeScriptClass = (function () { function MyTypeScriptClass() { } MyTypeScriptClass.prototype.foo1 = function () { alert('invoked foo1'); }; MyTypeScriptClass.prototype.foo2 = function () { alert('invoked foo2'); }; return MyTypeScriptClass; })(); ``` What I am looking for is typescript that can compile to the following JavaScript: ``` var fvm = new FlasherViewModel2(); var MyTypeScriptClass = (function () { function MyTypeScriptClass() { } MyTypeScriptClass.prototype.foo1 = function () { alert('invoked foo1'); }; MyTypeScriptClass.prototype.foo2 = function () { alert('invoked foo2'); }; function foo3() { alert('invoked foo3'); } return MyTypeScriptClass; })(); ``` Can it be done? (As a side note, I know that foo3 would not be callable from external code. I would actually invoke foo3 from another method within the class, for example, to pass a function to a jQuery fadeOut.)