How to call "super" from a callback in coffeescript

user_158285 0
Viewed 4349
class Foo
    a: ->
        x.call =>
            super

will not compile as I can't call super from anonymous class. However my intention is to call the superclass method for 'a'. Is this a missing capability in coffeescript?

Note that I change the code to

class Foo
    a: ->
        x.call =>
                return Foo.__super__.a.apply(this, arguments)

to make it work but that just ain't coffeescript right!

coffeescript super
3 Answers

I've encountered the same problem and solved it like this:

ChainSuper = (classType, memberName, useFatArrow) ->
  original = classType::[memberName]
  superf  = classType.__super__.constructor.prototype[memberName]
  callSuper = (thisRef, args, superArgs) ->
    for i in [0...superArgs.length]
      args[i] = superArgs[i]
    superf.apply thisRef, args

  classType::[memberName] = if useFatArrow
    (args...) ->
      original.call @, args..., (superArgs...) =>
        callSuper @, args, superArgs
  else
    (args...) ->
      original.call @, args..., (thisRef, superArgs...) ->
        callSuper thisRef, args, superArgs

This is very non-kosher and will probably break if the Coffeescript compiler's class semantics change. While waiting for that, however, I'm able to say:

CheckSuper = (ref, message) ->
  if ref instanceof Superclass
    alert "Works with #{message}"

class Superclass
  plus: (values...) ->
    CheckSuper @, 'plus'
    @val = 0
    for i in values
      @val += i
    alert @val

  minusBase: 0
  minus: (values...) ->
    CheckSuper @, 'minus'
    @val = @minusBase
    for i in values
      @val -= i
    alert @val

class Subclass extends Superclass
  plus: (values..., Super) ->
    setTimeout (=> Super @), 0
  ChainSuper @, 'plus'

  minus: (values..., Super) =>
    @minusBase = values[0]
    setTimeout (-> Super values[1...]..., 0), 0
  ChainSuper @, 'minus', true


subInstance = new Subclass()
subInstance.plus 1, 2, 3, 4
minus = subInstance.minus
minus 100, 10, 1

Please notice, that with this solution, if you pass n arguments to a Super() when the method has been called with more than n arguments, only the first n arguments get overwritten in the call to the super method and the rest will be passed on unchanged.

user_13129350
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