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Subject: Re: [boost] [type_erasure] default implementation of concept signatures
From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2011-05-28 12:52:21


On 05/28/2011 09:42 AM, Vicente Botet wrote:
> Concepts are able to define default implementations if the type doesn't
> provides the specific signature. For example in 20.1.2 Comparisons the
> concept LessThanComparable defines a default for implementation for
> bool operator>(U const& a, T const& b) { return b < a; }
> bool operator<=(U const& a, T const& b) { return !(b < a); }
> bool operator>=(T const& a, U const& b) { return !(a < b); }
> In the definition of the concept less_than_comparable in Boost.Any, you
> define the implementation of these operators in function of operator<(), but
> if the types provide the specific signature the specific function is not
> called. I guess that the default implementation of for example operator<=
> should use operator< only if the the underlying types don't provide
> operator<=.
> What do you think?

I don't think this is worth the extra
complexity. Do you know of any real
case where it will either
a) change the behavior of a program or,
b) make the program more efficient?

It will definitely make the vtable

In Christ,
Steven Watanabe

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