[Moderator's note: this defect report has been
 forwarded to the C++ committee. -moderator.]

Section: 14.8.2.4 - Deducing template arguments from a type
[temp.deduct.type]
Submitter: Andrei Iltchenko (iltchenko@yahoo.com)

Paragraph 9 of the section 14.8.2.4 [temp.deduct.type] enumerates the
forms that the types P and A need to have in order for template
argument deduction to succeed.

For P denoting a pointer to function the paragraph lists the following
forms as allowing for template argument deduction:
type(*)(T)
T(*)()
T(*)(T)

On the other hand, no provision has been made to accommodate similar
cases for references to funtions, which in light of the wording of
14.8.2.4 [temp.deduct.type] paragraph 11 means that the program below
is ill-formed (some of the C++ compilers do not reject it however):


template<typename Arg, typename Result, typename T>
Result  foo_r(Result(& rf)(Arg), T x)
{   return  rf(Arg(x));   }

template<typename Arg, typename Result, typename T>
Result  foo_p(Result(* pf)(Arg), T x)
{   return  pf(Arg(x));   }

#include <iostream>
int  show_arg(char c)
{
   std::cout << c << ' ';
   if(std::cout)  return  0;
   return  -1;
}


int  main()
{
                                               // The deduction
   int  (& rf1)(int(&)(char), double) = foo_r; // shall fail here

                                               // While here
   int  (& rf2)(int(*)(char), double) = foo_p; // it shall succeed

   return  rf2(show_arg, 2);

}


Suggested resolution:

In the list of allowable forms for the types P and A (paragraph 9 of
the section 14.8.2.4 [temp.deduct.type]) replace

type(*)(T)
T(*)()
T(*)(T)

by

type(T)
T()
T(T)

Given the latitude provided by paragraph 10 of the section 14.8.2.4
[temp.deduct.type], this will make it possible for template argument
deduction to be done against pointers and references to functions
alike.


Regards,

Andrei Iltchenko.
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