Re: Explicitly specializing std::min() on VC++ 2005 Express Edition

From:
"Tom Widmer [VC++ MVP]" <tom_usenet@hotmail.com>
Newsgroups:
microsoft.public.vc.language
Date:
Tue, 08 May 2007 11:30:02 +0100
Message-ID:
<uCq1XvVkHHA.4188@TK2MSFTNGP02.phx.gbl>
Matthias Hofmann wrote:

"Tom Widmer [VC++ MVP]" <tom_usenet@hotmail.com> schrieb im Newsbeitrag
news:eyKHNIkjHHA.4872@TK2MSFTNGP03.phx.gbl...

template <std::size_t N>
struct min_impl<char[N]>;

This specialization for non-const char[N] seems superflous, as
minimum<T>() accepts a constant reference, so min_impl<T>::impl() will
never see a non-const char[N].

Good point, though I think it's the other way around - char[N] is needed,
but not char const[N].


When I spezialize as follows:

// Non-const version.
template <std::size_t N>
struct minimum_impl<char[N]>;

then I get an error when calling minimum_impl::impl<>() from within
minimum<>() because the actual parameters are constant.


Interesting, I just tried that on VC2005 and got an error too (the same
error?). The error is due to a compiler bug, but there's a very simple
workaround, which is to drop the use of ?:.

template <std::size_t N>
struct minimum_impl<char[N]>
{
  static char const (&impl(char const (&a)[N], char const (&b)[N]))[N]
   { if (std::strcmp(a, b) < 0)
      return a;
   return b; }
};

The problem is that VC2005 is ignoring 5.16/4 (about ?: ):
"If the second and third operands are lvalues and have the same type,
the result is of that type and is an lvalue."

Instead, it is applying the lvalue-to-rvalue conversion (e.g. the
array-to-pointer conversion) to the parts of the ?: expression, and thus
trying to return a pointer where an array is expected.

On the other hand,

when I spezialize as follows:

// Const version.
template <std::size_t N>
struct minimum_impl<const char[N]>;

then the primary template is called, not the specialization. I found out
that this is because there is a bug in your implementation of minimum<>().
It should look like this:

template <class T> inline
const T& minimum( const T& a, const T& b )
{
    // Note the 'const' before 'T'.
    return minimum_impl<const T>::impl( a, b );
}


That's a bit of an odd way of doing it, since it means you always have
to specialize for const T, rather than T, as is usual.

C++ declaration syntax isn't very nice.


Indeed, that's why I am trying a typedef:

template <std::size_t N>
struct minimum_impl<const char[N]>
{
    typedef const char( &ARG )[N];

    static ARG impl( ARG a, ARG b )
    {
        // C2440 error here.
        return std::strcmp( a, b ) < 0 ? a : b;
    }
};

That looks much clearer, but on VC++ 2005 Express Edition, it does not work.
I get a C2440 error for not being able to convert from 'const char*' to
'const char (&)[2]' in the return statement. Where does that 'const char*'
come from?


See above. Just to be clear, this compiles and works on VC2005:

#include <cstring>

template <class T>
struct minimum_impl
{
   static const T& impl(const T& a, const T& b);
// { return a < b ? a : b; } commented out to check correct one is
//being called.
};

template <std::size_t N>
struct minimum_impl<char[N]>
{
  static char const (&impl(char const (&a)[N], char const (&b)[N]))[N]
   { if (std::strcmp(a, b) < 0)
      return a;
   return b; }
};

template <class T> inline
const T& minimum( const T& a, const T& b )
{
     // Note no 'const' before 'T'.
     return minimum_impl<T>::impl( a, b );
}

int main()
{
    char foo[] = "hello";
    char bar[] = "world";

    minimum(foo, bar);
}

Tom

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