What should `super` refer to?

[lydell]

I know that super has been discussed many times before (#2638 and #3436 for example), but in my opinion this is something new. This issue is not about changing super totally, but merely about defining its edge cases. The reason I started this discussion is because of issues I bumped into when making #3785.

This issue is not about what super without parentheses should do etc. It is only about what super should refer to in different cases.

The only way I’ve ever used super, and the only way I’ve even seen super be used, is like this:

class A extends B
  method: -> super

That is, only in an instance method defined using the method: syntax inside a class body with extends.

If that was the only case super was allowed, compiling it would be trivial. The above could compile to:

var A;

A = (function(_super) {
  __extends(A, _super);

  function A() {
    return _super.apply(this, arguments);
  }

  A.prototype.method = function() {
    return _super.prototype.method.apply(this, arguments);
  };

})(B);

CoffeeScript also allows super in “static” methods:

class A extends B
  @method: -> super

That would also be easy to compile:

var A;

A = (function(_super) {
  __extends(A, _super);

  function A() {
    return _super.apply(this, arguments);
  }

  A.method = function() {
    return _super.method.apply(this, arguments);
  };

})(B);

If dynamic keys are implemented, that’d be straightforward too:

class A extends B
  "#{foo()}": -> super

var A;

A = (function(_super) {
  var _name;

  __extends(A, _super);

  function A() {
    return _super.apply(this, arguments);
  }

  A.prototype[_name = "" + foo()] = function() {
    return _super.prototype[_name].apply(this, arguments);
  };

})(B);

So far so good. All nice and simple.

CoffeeScript also allows a “classic” style for making “classes”:

A = ->
A extends B
A::method = -> super

Note that the above isn’t equivalent to the class examples, since the constructor A doesn’t apply B. That could be done by writing A = -> B.apply this, arguments manually, but it would be much nicer to be able to write A = -> super, but that throws an error.

There is no way to make a static method call super using the “classic” style. A.method = -> super throws an error.

So when is it possible to use super “classic” style?

A = -> super # throws
A.m = -> super # throws
A::m = -> super # works
A.prototype.m = -> super # works
A["prototype"].m = -> super # throws
A["pro" + "totype"].m = -> super #throws
A[prototype].m = -> super #throws

According to #1392 you should be able to namespace classes (properly implemented in #3785):

namespaced.A::m = -> super # works

So what’s the problem? First of all, we don’t know the superclass, unlike in the class examples above. Therefore, the superclass is assumed to be accessible at A.__super__.

• A: Could be compiled to use A.__super__.constructor.

• A.m: Could be compiled to use A.__super__.constructor.m.

• A::m, A.prototype.m: No problems. If we allow the above though, it might be confusing that super refers to different things in A.Prototype.m and A.prototype.m.

• A["prototype"].m: If A.prototype.m is allowed, shouldn’t this also be? Ok, we could do it.

• A["pro" + "type"].m: If A["prototype"].m is allowed, shouldn’t this also be? Ough, I see where this is going.

• A[prototype].m: If A["prototype"].m is allowed, I’d expect this to be allowed, too. However, we don’t know if prototype is "prototype". A[f()].m=->super could compile to:

  var _super, _ref;
  
  _super = (_ref = f()) === "prototype" ? A.__super__ : A.__super__.constructor;
  A[_ref].m = function() {
    _super.apply(this, arguments);
  };

  But perhaps that single-time runtime check is a bit weird. Also it feels odd that A[f()].m=->super does different things to super depending on the value of f(). Therefore it might be better to compile it to assume that it always means a definition of a static method. I mean, why would you ever use something else than A::m or A.prototype.m if you really wanted to create an instance method? There’s no reason to.

  But even if A[b].m always meant static method, should A["prototype"].m too? What about A["pro" + "type"].m?

There’s also the case where super is used in an object literal:

extend A,
  method: -> super

However, why use an extend library function when you’re using CoffeeScript and have extends? And shouldn’t that extend function also provide some alternative to super? So I guess this case should be disallowed, just like [1, (-> super), 2] is.

Summing up, this is what I think:

The A.__super__ thing is ugly.

super is not “classic”: It does not belong to the “classic” style of making classes. If you’re using the classic style you’re not using super—instead, you might use some manual/“classic” way of doing it.

So in my opinion, super should only be allowed in a function after a method: or @method: in a class body. Nowhere else. That’s where they belong, and that’s where they’re simple. That’s also the only place I’ve ever seen them used. Compile it like the first two examples. Keep it simple! And as far as I understand, super will only be available in class blocks in ES6 (I might be wrong, though).

If that proposal is not acceptable, I’d say that super should refer to:

A = -> super
  → A.__super__.constructor
A.method = -> super
  → A.__super__.constructor.method
A.prototype = -> super
  → A.__super__.constructor.prototype
A.prototype.method = -> super
  → A.__super__.method 
A.Prototype.method = -> super
  → A.Prototype.__super__.constructor.method
A[anything].method = -> super
  → A[anything].__super__.constructor.method
A["prototype"].method = -> super
  → A["prototype"].__super__.constructor.method
  A known “gotcha”. `[]` notation always means static method. CoffeeScript should not try to interpret the code inside the brackets.

Oh, and if anyone brings up ES6 super, as far as I understand super() calls the method with the same name of the superclass, while super.anySuperMethod() calls “anySuperMethod” of the superclass.

What are people’s thoughts?

[epidemian]

The A.__super__ thing is ugly.

super is not “classic”: It does not belong to the “classic” style of making classes. If you’re using the classic style you’re not using super—instead, you might use some manual/“classic” way of doing it.

Totally agreed.

I also find it very inelegant that A::method = -> super or A.protoype.method = -> super work, while other seemingly equivalent forms don't:

# Extracting the prototype into a variable doesn't work:
proto = A.prototype
proto.foo = -> super # Error

# Nor does using a function to assign the methods to the prototype:
_.extend A.prototype,
  method: -> super # Error

---

That being said, however, couldn't we make expr().method = -> super or expr1()[expr2()] = -> super work in some way? Would compiling to something like this work?:

// For `expr().method = -> super`
var _ref, _super;
_super = (_ref = expr()).method
_ref.method = function() {
  _super.apply(this, arguments)
};

// For the more general `expr1()[expr2()] = -> super` case:
var _ref1, _ref2, _super;
_super = (_ref1 = expr1())[_ref2 = expr2()];
_ref1[_ref2] = function() {
  _super.apply(this, arguments)
};

[lydell]

That will probably raise a"TypeError: Cannot call method 'apply' of undefined", wouldn't it?

Edit: Just checked, it does.

[epidemian]

@lydell, this is a little example of what i meant: http://jsfiddle.net/18nqo06h/

// class A
//   foo: -> console.log 42
var A = function() {};
A.prototype.foo = function() { console.log(42); }

// class B extends A
var B = function() {};
B.prototype = new A;

var f = function() { return B.prototype; };
var g = function() { return 'foo'; };

// f()[g()] = -> super
var _ref1, _ref2, _super;
_super = (_ref1 = f())[_ref2 = g()];
_ref1[_ref2] = function() {
  return _super.apply(this, arguments)
};

new B().foo() // 42

It has its caveats though. For instance, if A::foo was not defined at the time B::foo is defined, it would break; or if A::foo was changed after B::foo was defined, then B::foo would still call the original version of A::foo when doing super. But maybe these could be reasonable/expectable caveats of calling super outside a class. What do you say?

[lydell]

Now I understand. Clever!

[epidemian]

I now realize that it is still a very limited solution:

# Assigning the super-calling function to a variable wouldn't work:
f = -> super
A::method = f

# Nor would using a super-calling function as a method on a literal object: 
_.extend A.prototype, 
  method: -> super

So, yes, i would vote in favour of limiting the super syntax to methods inside a class statement.

Does ES6 allow super calls outside classes? On standalone functions or in object literals?

Does this "work" in ES6?

var a = {
  foo() { super() }
};

Object.setPrototypeOf(a, {
  foo() { console.log(42); }
});

a.foo() // 42?

(It does not compile on 6to5.org, but i don't know if that's per spec or not; something similar using a class does work)

[lydell]

The only way I’ve ever used super, and the only way I’ve even seen super be used, is like this:

class A extends B
  method: -> super

Is there anyone who is a pro at searching Github and could back up this? If so, the decision to limit super to this case would be easier.

[vendethiel]

From a quick search:

https://github.com/search?l=CoffeeScript&q=super+language%3ACoffeeScript&ref=advsearch&type=Code&utf8=%E2%9C%93

[lydell]

Looking through the first ten pages as well as a few of the last and a bunch in the middle of those search results, I found one occasion where super wasn't used like above, but that one was a mistake by the developer.

[jashkenas]

In general the current idea is to mostly limit super use to methods within class bodies — and indeed you note that this is how it is almost always used. We're never going to be able to completely statically analyze calls that are constructed in crazy ways, and we shouldn't try.

Can you fill me in on exactly what situations we would lose support for if we followed @lydell's "keep it simple" suggestion?

[lydell]

The following uses of super would be parse errors:

class A
  m: -> super # missing `extends`
  @m: -> super # ditto
  @::m = -> super # regular assignment not allowed
  @m = -> super # ditto

A::m = -> super # ditto
A.prototype.m = -> super # ditto

This is would continue to be allowed:

class A extends B
  m: -> super
  @m: -> super

Edit: Since #3785 has been merged the new cases allowed by that PR would become parse errors, too.

[jashkenas]

And that would allow us to remove the __super__ property entirely?

[lydell]

Yep (see the beginning of the first post).

[jashkenas]

I'm very tempted — so much so that I started typing out a "go for it" response before changing my mind — but I don't think that we should make that change. The main reason is this case:

class A extends B
  ... blah blah ...

... and then later, in another file, say a-extensions.coffee:

A::method ->
  super
  ... continue refining method ...

Removing the __super__ property would be more elegant, but it would prevent CoffeeScript classes from being "open" except in the same lexical scope where they're first defined. It's not the most killer restriction, but it's nice that you can still add methods to them outside of that scope, and call super like normal.