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From "Farid Zaripov" <>
Subject RE: [PATCH] dynatype.cpp example and MSVC
Date Fri, 30 Mar 2007 09:59:39 GMT
> -----Original Message-----
> From: Martin Sebor [] 
> Sent: Thursday, March 29, 2007 9:38 PM
> To:
> Subject: Re: [PATCH] dynatype.cpp example and MSVC
> >   The dynatype::operator= (const dynatype::map<void>&) 
> removed because it's never used.
> Could it ever be used? Imagine dynatype being a general class 
> provided by a library. Does there exist a valid use case for 
> this operator? If not, I have no problem with removing it.
> Otherwise, if it doesn't break anything, let's keep it.

  The dynatype::map<void> type is a private type. So that operator= can
be used
only within any member of the dynatype class. So that operator= should
be invoked
using following code:

*this = [some variable of dynatype::map<void> type];

  But that "*this = ..." code used only in two places:

1. in dynatype::dynatype<dynatype::map<void> > (const
dynatype::map<void> &t);
  that constructor cannot be invoked because of dynatype::map type is

2. in template <class T> void dynatype::copy (const dynatype &rhs); with
T = dynatype::map<void>
  but there exist an empty specialization for that T:

template <>
inline void dynatype::copy<dynatype::map<void> >(const dynatype&)
{ /* no-op */ }

  If we would replace the dynatype::map<void> specializations to the
void specializations,
we cannot keep the dynatype::operator= (const dynatype::map<void>&)
because of
that operator after the replacement would be dynatype::operator= (const
void&), but
that syntax is incorrect.


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