How to understand this piece of C++ code, which is for type checking？

Question

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How does the following code work?

#define TYPE_CHECK(T, S)                                       \
  while (false) {                                              \
    *(static_cast<T* volatile*>(0)) = static_cast<S*>(0);      \
  }

I found it tricky, how does it work? And why volatile?

@iammilind: it won't be executed, only compiled, that's the whole point — Karoly Horvath, Aug 08 '11 at 09:36

score 4 · Answer 1 · answered Aug 08 '11 at 09:33

4

Since it's a static cast, it's getting the compiler to ensure that the pointers to the two types are equivalent. The while(false) ensures it's never run and that it's only a compile time thing.

answered Aug 08 '11 at 09:33

Preet Sangha

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Ah, nice! +1. I would never thing about such thing. – Kiril Kirov Aug 08 '11 at 09:34
It would be nice if you could explain how exactly the code works. Especially what effect making the pointer volatile has. – sepp2k Aug 08 '11 at 09:36
I always have to remember rinterpret_cast => runtime – Preet Sangha Aug 08 '11 at 09:36
See this: http://stackoverflow.com/questions/5237163/how-does-the-following-code-work – Preet Sangha Aug 08 '11 at 09:37

score 1 · Answer 2 · answered Aug 08 '11 at 09:34

1

It statically checks that a S* can be cast into a T*. No code is executed at runtime. I think the volatile prevents optimization, which would just make the compiler ignore the code.

answered Aug 08 '11 at 09:34

slaphappy

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Optimization would not permit the compiler to ignore the code entirely, since the compiler still has to diagnose invalid code whether it's optimized away or not. I suspect the `volatile` is there to suppress a warning from some compiler that notices a null pointer is dereferenced. Well, it isn't since the expression isn't evaluated, but the expression says to do it. If I'm right, then `volatile` indicates "user appears to think he knows what he's doing, and really wants to dereference that null pointer, so don't warn". – Steve Jessop Aug 08 '11 at 10:23

score 0 · Answer 3 · answered Aug 08 '11 at 09:35

0

The code is checking if S* can be assigned to T*, and forcing a compile error if it cannot. As for volatile I'm not sure.

answered Aug 08 '11 at 09:35

john

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score -1 · Answer 4 · answered Aug 08 '11 at 09:35

-1

If the types can't be casted, it won't compile. But it is never executed.

answered Aug 08 '11 at 09:35

Victor

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How to understand this piece of C++ code, which is for type checking？

4 Answers4