Typescript: Infer static member references

[jez9999]

In the following code:

export default class SettingsParser {
    private static _settings: Settings;

    /** Parse commandline, check settings, and init logging to appropriate loglevel */
    public static parse(settings: Settings) {
        _settings = settings;
    }
}

The _settings reference is flagged as an error with the message "[ts] Cannot find name '_settings'. Did you mean the static member 'SettingsParser._settings'?" Yes, I did. C# just infers this. Why can't TypeScript? Having to write SettingsParser. each time is very verbose.

[andy-ms]

TypeScript just type-checks JavaScript code -- it doesn't alter it. Since your example won't work when run as JavaScript, TypeScript will issue an error.

[jez9999]

It alters it a bunch, like making import statements into require statements if necessary. Why not add the class name when it is useful? It doesn't work run as JS but that's not a design goal of TS - the design goal is "valid JavaScript is valid TypeScript", not the other way around.

[DanielRosenwasser]

See https://github.com/Microsoft/TypeScript/wiki/TypeScript-Design-Goals

Non-goal: Add or rely on run-time type information in programs, or emit different code based on the results of the type system. Instead, encourage programming patterns that do not require run-time metadata.

[kitsonk]

It alters it a bunch, like making import statements into require statements if necessary.

That is transforming one form of valid TypeScript to another form. Transpiling.

[jez9999]

That's somewhat splitting hairs. It's changing variable names. Fundamentally it is changing the input code. Why does it matter if it isn't valid JavaScript going in?

[jez9999]

@DanielRosenwasser So if it's a "bad idea" to emit this different code, what would one suggest as a good alternative? Do we really have to write out the whole class name every time?

[andy-ms]

You could just use a let in the scope of the class.

let _settings: Settings;
export default class SettingsParser {
    public static parse(settings: Settings) {
        _settings = settings;
    }
}

The reason we won't emit the class name for you is that that would change the semantics of JavaScript -- setting a member of a class and setting a local variable are two different actions with different syntax. Observe that in this example _settings = settings; does not set a static property of the class, it sets a local variable.
EDIT settings -> _settings

[jez9999]

@andy-ms You mean let _settings: Settings;? Is that logically the same as a static member, though?

[andy-ms]

It's not the same as a static member since it's not a property on the class (classes are objects and can have properties) -- it can only be accessed in its scope. But since you wanted the property to be private anyway, that's probably what you want.

[DanielRosenwasser]

If you need that, there's

let _settings = {};
export default class SettingsParser {
    get settings() {
        return _settings;
    }
    set settings(value: typeof _settings) {
        _settings = value;
    }
}

[mhegazy]

Automatically closing this issue for housekeeping purposes. The issue labels indicate that it is unactionable at the moment or has already been addressed.