#707 Equivalent of "super" for mixins?

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helium Wed 5 Aug 2009

mixin Foo {
   virtual Void method() { ... }
}

class Bar {
   override Void method() 
   { 
      ... 
      // how do I call Foo's "method"?
      // something like super.method() but for a mixnin not the super class
      ...
   }
}

brian Wed 5 Aug 2009

Use a named super like Foo.super.method.

See docLang.

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