Re: Why doesn't implicit conversion work with wide ostream?

Question @ SO:
http://stackoverflow.com/questions/17486156/why-doesnt-implicit-conversion-work-with-wide-ostream

I have some behaviour that I do not understand. I observed this on
VS2005, but [IDEONE (using GCC 4.7.2) outputs][http://ideone.com/doSHvq]
basically the same.

Here's the code:

     #include <iostream>
     #include <string>

     struct UserString {
         const char* p;
             operator const char*() const {
                     std::cout << "! " << __FUNCTION__ << std::endl;
                     return p;
             }

             UserString()
             : p ("UserString")
             { }
     };

     struct WUserString {
             const wchar_t* p;
             operator const wchar_t*() const {
                     std::cout << "! " << __FUNCTION__ << std::endl;
                     return p;
             }

             WUserString()
             : p (L"WUserString")
             { }
     };

     int main() {
             using namespace std;
             cout << "String Literal" << endl;
             cout << string("std::string") << endl;
             cout << UserString() << endl;
             cout << static_cast<const char*>(UserString()) << endl;

             wcout << L"WString Literal" << endl;
             wcout << wstring(L"std::wstring") << endl;
             wcout << WUserString() << endl;
             wcout << static_cast<const wchar_t*>(WUserString()) << endl;

             return 0;
     }

Here's the output:

     String Literal
     std::string
     ! operator const char* **** "works"
     UserString ****
     ! operator const char*
     UserString
     WString Literal
     std::wstring
     ! operator const wchar_t* **** "doesn't" - op<<(void*) is used
     0x80491b0 ****
     ! operator const wchar_t*
     WUserString

What's going on here?!?