Printing CComBSTR with iostream (std::wcout)

Question

The following code:

#include <iostream>
#include <atlstr.h>

int main()
{
    CComBSTR bstr(L"test");
    std::wcout << bstr << std::endl;
    std::wcout << static_cast<BSTR>(bstr) << std::endl;
}

prints

033FA16C
test

I tried to investigate which conversions take place in each case with a debugger but both time had stepped into operator BSTR. So why the first line prints address while the second prints a text?

Answer 1

We can remove ATL from this entirely, as it's really a question of how wcout works.

Consider the following minimal example:

#include <iostream>

struct Foo
{
    operator const wchar_t*() const { return L"what"; };
};

int main()
{
    Foo f;
    std::wcout << f << std::endl;
    std::wcout << (const wchar_t*)f << std::endl;
}

// Output:
//   0x400934
//   what

(live demo)

In your example, the implicit conversion from CComBSTR to BSTR is triggered, but not by the template that instantiates operator<<(const wchar_t*) (because the conversion is "user-defined", and user-defined conversions are not considered for template parameter matching). The only viable candidate then is the non-template operator<<(const void*), to which your converted BSTR is passed.

There's actually a proposal to "fix" this in the standard (LWG 2342) and the text of the proposal explains this in more detail.

In summary:

For wide streams argument types wchar_t const* and wchar_t are supported only as template parameters. User defined conversions are not considered for template parameter matching. Hence inappropriate overloads of operator<< are selected when an implicit conversion is required for the argument, which is inconsistent with the behavior for char const* and char, is unexpected, and is a useless result.

The only remaining viable overload is the one taking const void* and, since every pointer can implicitly convert to const void*, that's what you're getting.