I believe they are doing code analysis, at least that's what the README for Roll...

frutiger · on March 22, 2016

I don't think that's possible in all cases:

    var mod = {};
    module.foo = function () { return 42; };
    module.bar = function () { return -1; };

    module[Math.random() > 0.5 ? "foo" : "bar"]();

How can `Rollup` determine that I need both functions here?

AnkhMorporkian · on March 22, 2016

That's why it only works with ES6 imports. ES6 imports must be made at the top of the file, and aren't dynamic. You must explicitly import what you need.

Additionally, this method only works with named imports. So for example, you can have two files:

File a.js

    export a = 5;
    export b = 6;
    export c = 7;
    export d = 8;
    export default {a, b, c, d};

File b.js

    import {a, b} from "./a.js";

    console.log(a);
    console.log(b);

c/d will never be included in the output file, nor will the default export. However, if you do:

    import a from "./a.js";

Then it won't optimize at all, because all of those objects are referenced in the default export.

Note that it's possible I'm wrong as to the specific optimization method, but largely speaking this is how it should work.

frutiger · on March 23, 2016

That much makes sense, but if I understand GGGP's example properly, they are talking about pruning unused internal references within a module, but GGP's response implies that somehow Rollup will prune those out.

AnkhMorporkian · on March 23, 2016

Ah, I understand the confusion now. I'm not 100% sure how that code works, but I do know it uses estree and that case is tested for, nearly verbatim. Here's a link to the source which handles that sort of stuff.

https://github.com/rollup/rollup/blob/6fc8631ba1aad718885151...