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From: Michael Fawcett (michael.fawcett_at_[hidden])
Date: 2007-10-09 13:15:27

On 10/9/07, Dean Michael Berris <mikhailberis_at_[hidden]> wrote:
> I thought about this, and the only problem is that it assumes that
> 'include' does not replace the already existing mapping between a
> function signature and the actual function being 'included'. Consider:
> void foo(int) { };
> void bar(int) { };
> overloads<void(int), void(std::string)> functions;
> functions.include(&foo);
> functions.include(&bar);
> functions(1); // will imply that 'foo' and 'bar' will both be called
> The above code suggests that functions(1) should call both foo and
> bar, because both functions are included in the overload set.

It doesn't suggest that to me. If we were talking normal overload
sets, it would fail to compile because it was ambiguous. The user is
trying to introduce an ambiguity with the second include line, and
he's ignoring the return value, which I presume contains 'false', as
in, not inserted.

--Michael Fawcett

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