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From: Andreas Huber (andreas_at_[hidden])
Date: 2001-11-21 17:36:23

Hi Peter

"Foreign" refers to a function object the class of which is not part
of the boost library (and not an instantiation of a template provided
by boost). As outlined in my previous post:

// MyFunctor is "foreign" ...
class MyFunctor
    void operator()();

int main()
  // ... f1 is not
  boost::function< void > f1 = MyFunctor();

I see that with the design of boost::function ANY function object is
as good as another. However, that would not be possible if
boost::function provided equality operations. Please see my reply to



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