[Bug c++/15882] New: Check for return type of overloaded operator new too early

[bangerth at dealii dot org]

I don't know what to say about this, but I'd like to solicit second 
opinions -- this fails: 
----------------------- 
#include <new> 
 
template <bool C, typename T> struct SFINAE; 
template <typename T> struct SFINAE<true,T>  
{ 
    typedef T type; 
}; 
 
SFINAE<true, void *>::type 
operator new (size_t, int); 
 
template <typename T> 
typename SFINAE<T::condition, void *>::type 
operator new (size_t, T); 
------------------------ 
 
g/x> /home/bangerth/bin/gcc-3.5-pre/bin/c++ -c y.cc 
y.cc:14: error: `operator new' must return type `void*' 
 
I believe that the check for the return type is set too early, while 
parsing the declaration of the template, not during instantiation. As 
is shown in the first overload of the operator, gcc is quite happy 
with a somewhat convoluted way to specify a return type of 'void*'. 
In the second overload, this may still be possible, if just the type 
'T' has an appropriately defined static member condition, or something 
else that can be converted to a bool evaluating to 'true'. However, the 
fact that we already get the error message before we even try to  
instantiate something seems wrong. 
 
W.

-- 
           Summary: Check for return type of overloaded operator new too
                    early
           Product: gcc
           Version: 3.5.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P2
         Component: c++
        AssignedTo: unassigned at gcc dot gnu dot org
        ReportedBy: bangerth at dealii dot org
                CC: gcc-bugs at gcc dot gnu dot org


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=15882