What should `new Loader({/* no hooks */});` do?

[unscriptable]

If a dev constructs a new loader with no hooks (e.g. new Loader({/* no hooks */});), will the loader be functional? Does it take-on some default behavior based on the environment? Or does it throw an exception if you try to import using it?

[juandopazo]

The way it is spec'd right now, hooks override methods in the new Loader instance. So these two are equivalent:

var loader = new Loader({ fetch: fn });

var loader2 = new Loader();
loader2.fetch = fn;

This means that it'll default to Loader.prototype.fetch (which throws). From 26.3.1.1.9:

If hook is not undefined, If isCallable(hook) is false, throw a TypeError exception. Let result be CreatePropertyOrThrow(loader, name, hook).

See http://people.mozilla.org/~jorendorff/es6-draft.html#sec-%loader%.

So yes, it throws.

[johnjbarton]

The environment default behavior is baked into the System instance of Loader, not into the class Loader.

[unscriptable]

thanks guys!