[Bug c++/55002] New: trailing return type is rejected in function signature

[Bug c++/55002] New: trailing return type is rejected in function signature

[leonid at volnitsky dot com]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=55002

             Bug #: 55002
           Summary: trailing return type is rejected in function signature
    Classification: Unclassified
           Product: gcc
           Version: 4.8.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: c++
        AssignedTo: unassigned@gcc.gnu.org
        ReportedBy: leonid@volnitsky.com


7.1.6.4 auto specifier [dcl.spec.auto]
note 2:  The auto type-specifier may appear with a function declarator
with a trailing-return-type (8.3.5) in any context where such a
declarator is valid 

Contrived example:
-------------------------------------------------------------
typedef unsigned long size_t;

// overload c-array
        template<class T, size_t N>
        size_t
size(T (&)[N]) { return N; }

// overload - c-string 
        template<size_t N>
        size_t
size(char (&A)[N]) { size_t i=0;  while(A[i++]);  return i; }

// APPLY

// this works
template<class T, size_t N> size_t   apply (const T (&A)[N], size_t (*f)       
 (const T (&)[N]))

// this rejected -- error: parameter declared ‘auto’
//template<class T, size_t N> size_t   apply (const T (&A)[N], auto (*f) ->
size_t (const T (&)[N]))

{ return   f(A); };

int main() {
        int A[2];
        return apply(A, size);
}
------------------------------------------------------------------

Without trailing return type in signature it is impossible to write apply-like
function with can deduce function-type for overloaded functions.

[Bug c++/55002] trailing return type is rejected in function signature

[leonid at volnitsky dot com]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=55002

--- Comment #1 from Leonid Volnitsky <leonid at volnitsky dot com> 2012-10-20 19:20:00 UTC ---
I've probably overcomplicated my example. Simpler test case:
----------------------------------
int f(auto (*ff) -> int (int) ) {
        return ff(1);
}
----------------------------------

[Bug c++/55002] trailing return type is rejected in function signature

[daniel.kruegler at googlemail dot com]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=55002

Daniel Krügler <daniel.kruegler at googlemail dot com> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
                 CC|                            |daniel.kruegler at
                   |                            |googlemail dot com

--- Comment #2 from Daniel Krügler <daniel.kruegler at googlemail dot com> 2012-10-20 20:24:19 UTC ---
Your issue looks invalid to me, because you are never providing a pointer to
function declarator. 

Lets look first at your simplified example:

auto (*ff) -> int (int)

This would be an invalid function declarator, because there is (a) no parameter
list and (b) the return type is a function type int(int), which is never
possible. I can only assume you mean

int f(auto (*ff)(int) -> int) {
  return ff(1);
}

and this is accepted. 

A similar defect is in your original code, where the only valid form would be

template<class T, size_t N> size_t   
apply (const T (&A)[N], auto (*f)(const T (&)[N]) -> size_t)
{ return   f(A); }


which is also accepted.

[Bug c++/55002] trailing return type is rejected in function signature

[paolo.carlini at oracle dot com]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=55002

Paolo Carlini <paolo.carlini at oracle dot com> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|UNCONFIRMED                 |RESOLVED
         Resolution|                            |INVALID

--- Comment #3 from Paolo Carlini <paolo.carlini at oracle dot com> 2012-10-20 22:55:59 UTC ---
Closing.