How well does MFC IPTR/CIP support class template argument deduction (CTAD)?

Raymond Chen

Continuing our investigation of which C++ COM wrappers support class template argument deduction (CTAD), next up is MFC’s IPTR/CIP.

The CIP template class fails CTAD because of these constructors:

template<class _Interface, const IID* _IID>
class CIP : public _CIP<_Interface, _IID>
{
public:
    typedef _CIP<_Interface, _IID> BC;
    typedef _Interface Interface;

    CIP(Interface* pInterface); // this one
    CIP(Interface* pInterface, BOOL bAddRef); // and this one

    ⟦ other stuff ⟧
};

In this case, the compiler can deduce the _Interface template parameter, but it doesn’t know what _IID to use. Again, adding some deduction guides can fix that by saying that it should use the __uuidof(T) as the _IID.

template<typename T> CIP(T*)       -> CIP<T, &__uuidof(T)>;
template<typename T> CIP(T*, BOOL) -> CIP<T, &__uuidof(T)>;

As noted before, injecting your own deduction guides into somebody else’s types is not recommended because it may conflict with deduction guides added by the type owner in the future.

Fortunately, in the case of MFC IPTR/CIP, you rarely use CIP directly. Instead, you use the IPTR that was generated from the CIP, and that type already knows what interface it is for, so there’s no need for CTAD.

using TestPtr = IPTR(Test);

void sample(Test* p)
{
    auto smart = TestPtr(p);
}

Okay, but what if you still want to create an MFC smart pointer from a raw pointer without having to specify the interface?

You can do what C++ did before CTAD was invented: “You kids with your CTADs and deduction guides. In my day, we had to use maker template functions,¹ and we liked it!”

template<typename T>
CIP<T, &__uuidof(T)> make_CIP(T* p)
{
    return p;
}

template<typename T>
CIP<T, &__uuidof(T)> make_CIP(T* p, BOOL bAddRef)
{
    return { p, bAddRef };
}

void sample(Test* p)
{
    auto smart1 = make_CIP(p);
    auto smart2 = make_CIP(p, FALSE);
}

This trick also works for _com_ptr_t, which is something I teased last time.

template<typename T>
_com_ptr_t<_com_IIID<T, &__uuidof(T)>>
    make_com_ptr_t(T* p)
{
    return p;
}

void sample(Test* p)
{
    auto smart = make_com_ptr_t(p);
}

¹ There are maker template functions all over the C++ standard library. std::back_inserter, std::make_reverse_iterator, std::make_pair, etc.

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  • 紅樓鍮 · Edited 0

    Bonus chatter: std::unique_ptr does not, and will never have deduction guides; that is,

    
    Foo *foo = bar();
    // this will never compile, even in C++77
    std::unique_ptr baz(foo);

    will never be well-formed. This is because baz could equally reasonably be std::unique_ptr<Foo> or std::unique_ptr<Foo[]>. (Unless C++77 gets, say, unification-based type inference…)

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