Is operator void*() conversion still part of C++ library?

Question

Consider this program:

#include <iostream>
int main()
{
    delete std::cout;
}

AFAIK the conversion function operator void* () const has been removed from C++11. So, this program should fail in compilation on a C++11 compiler. Ya, its true that both g++ 4.8.1 & 4.9.2 gives diagnosis (in the form of warning that deleting void* is undefined & that's also the good thing). But shouldn't this program fail in compilation because removal of that conversion function due to which all stream object could be implicitly converted to void* in C++98 & C++03?. Is this bug? It seems bit surprising that they still not have implemented this change.

I've tried this program in g++ 4.9.2(that supports C++14) but it gives warning not compiler error. Ideone compiler gives me an error as expected. (See live demo here)

Answer 1

It has nothing to do with the compiler, its a library issue. libstdc++ has lots of incompatibilities with C++11, of which this is just one. They are making breaking changes in 5 and up though iirc.

In short, it's neither a bug nor a compiler issue.