TypeScript does not set function name for MethodDeclarations

[azz]

This issue has been asked as a question on StackOverflow.

I have a TypeScript class, and I'm doing some metaprogramming where I need to be able to access instance.func.name, however TypeScript omits the function name in the compiled JS.

TypeScript:

class ClassName {
    // ... 
    func(): ReturnType {
        // ...
    }
}

Compiled JavaScript:

// ...
ClassName.prototype.func = function () {
    // ... 
};

Desired JavaScript:

ClassName.prototype.func = function func() {
    // ...                          ^^^^
};

Not sure if this is a bug or by design, but the source of the result is emitter.ts.

A proposed solution on StackOverflow was:

// src/compiler/emitter.ts:4671
function shouldEmitFunctionName(node) {
    if (node.kind === 173 /* FunctionExpression */) {
        // Emit name if one is present
        return !!node.name;
    }
    if (node.kind === 213 /* FunctionDeclaration */) {
        // Emit name if one is present, or emit generated name in down-level case (for export default case)
        return !!node.name || languageVersion < 2 /* ES6 */;
    }

    // MODIFIED
    if (node.kind === 143 /* MethodDeclaration */) {                    
        return true;
    }                                                        
}

[RyanCavanaugh]

We need to do this for ES6 compat

[RyanCavanaugh]

Open question -- what should happen when the function name is a quoted name that is an invalid identifier?

class Thing {
  'foo bar'() { }
}

var x = new Thing();
console.log(x['foo bar'].name);

ES6 prints foo bar, Babel prints fooBar (wat), TypeScript prints undefined.

[mhegazy]

set it explicitly?...

Object.defineProperty(Thing.prototype["foo bar"], "name", { configurable: true, writable: false, enumerable: false, value:"foo bar" });

[DanielRosenwasser]

You'd basically have to do the same thing we do for computed properties in object literals and split off at the first property with a string name whose name is not a valid identifier name.

[jeffreymorlan]

Another time using a named function wouldn't work is when the method accesses an outer variable with the same name:

var func;
class ClassName {
    func() { return func; }
}

Edit: a named function could also be incorrectly fetched by eval:

var func;
class ClassName {
    func(s) { return eval(s); }
}
new ClassName().func('func');

[MartyIX]

@jeffreymorlan has a point.

var func = "aa";
class ClassName {
    func() { return func; }
}
new ClassName().func();

returns "aa" in Chrome console

var func = "aa";
var ClassName = (function () {
    function ClassName() {
    }
    ClassName.prototype.func = function func() { return func; };
    return ClassName;
})();
new ClassName().func();

returns function func() { return func; } in Chrome console.