In C++/WinRT, what happens when I treat an IInspectable as or convert one to a bool
Last time, we looked at weirdness in how C++/CX treats hat pointers in a bool context. Fortunately, C++/WinRT is much less weird.
The IInspectable type supports a conversion to bool which tests whether the underlying pointer is null. It also supports comparison against nullptr which tests the same thing. And, unlike C++/CX, C++/WinRT uses this conversion for both explicit and contextual conversions.
IInspectable p = winrt::box_value(false); IInspectable q = winrt::box_value(false); if (p) std::cout << 1; if ((bool)p) std::cout << 2; if (static_cast<bool>(p)) std::cout << 3; if (p == q) std::cout << 4; if (p == false) std::cout << 5; if (!p) std::cout << 6; if ((bool)p == (bool)q) std::cout << 7;
| Condition | What’s happening | Result |
|---|---|---|
| if (p) | Tests p against nullptr. | prints 1 |
| if ((bool)p) | Tests p against nullptr. | prints 2 |
| if (static_cast<bool>(p)) | Tests p against nullptr. | prints 3 |
| if (p == q) | Compares two objects for identity. | does not print |
| if (p == false) | Not allowed (compiler error). | |
| if (!p) | Tests p against nullptr. | does not print |
| if ((bool)p == (bool)q) | Tests p and q against nullptr. | prints 7 |
Note that the last case prints 7 but not for the reason you think. It’s not doing any unboxing at all. It’s just checking whether both variables are non-null.
IInspectable t = winrt::box_value(true); if ((bool)p == (bool)t) std::cout << 8; // prints 8!
Bonus chatter: There is a little quirk in the p == false case. My understanding is that prior to C++11, false was a legal null pointer constant, but the rules in C++11 were tightened so that false is no longer a null pointer context.
Microsoft’s Visual Studio C++ compiler, however, continues to accept false as a null pointer constant, even in non-permissive mode. This means that if you’re using Microsoft’s Visual Studio C++ compiler, the fifth row of the table is slightly different:
| Condition | What’s happening | Result |
|---|---|---|
| if (p == false) | false converted to IInspectable{ nullptr } and compared with p |
does not print |
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4 comments
It looks like that MSVC behaviour has just been fixed, for the /permissive- case in the latest version.
https://developercommunity.visualstudio.com/content/problem/46351/integer-constant-expressions-implicitly-convert-in.html
Hello Raymond,
Micro-nitpick: Should you not initialize IInspectable p to true instead of false?
Fantastic blog – thank you so much!!
Thomas
p is initialized to false on purpose.
pis presumably initialized tofalseto highlight the fact, thatoperator==compares for identity, not equality.