Cast void pointer to int

Question

With a VC++ project I have a line:

FXint realIndex = (int)m_cmbDevice->getItemData(indx);

This reads a void pointer value from combo box to an int value. In VC++ it compiles and works well. Now I have to port this to Linux and there I get a compile error

cpp:514:20: error: 
      cast from pointer to smaller type 'int' loses information
        FXint realIndex = (int)m_cmbDevice->getItemData(indx);

Now I use

std::size_t x = reinterpret_cast<std::size_t>(m_cmbDevice->getItemData(indx));
FXint realIndex = x;

My question is now, is this the right way to go?

Yeah, that won't lose any info. But if `sizeof(FXint) < sizeof(size_t)`, then you will lose info at that point instead (i.e., in the actual assignment). — goodvibration, Nov 30 '20 at 08:38
Add a `static_assert(sizeof(FXint) == sizeof(void*), "Size of FXint doesn't match size of a pointer!");` before doing the cast, to make sure you won't lose any information — Kaldrr, Nov 30 '20 at 08:48
please show a [mre], what is returned by `m_cmbDevice->getItemData`? — Alan Birtles, Nov 30 '20 at 08:49
static_cast(yourPointer) If you want to use it to save a pointer I would recommend using a unsigned long instead of an int — just a guy, Dec 02 '20 at 12:55

phuclv · Answer 1 · 2020-11-30T09:44:38.630

1

~~reinterpret_cast is dangerous and is typically used only to interpret a pointer as another kind of pointer. You should use static_cast instead~~
size_t is not necessarily big enough to store a pointer. There are already intptr_t and uintptr_t which are signed and unsigned integer types that are capable of holding a pointer
FXint realIndex = x; still raises a warning if x is wider than FXint, so another cast is necessary to turn off all related warnings

So you need to do this

auto x = reinterpret_cast<uintptr_t>(m_cmbDevice->getItemData(indx));
auto realIndex = (FXint)x;

edited Nov 30 '20 at 09:44

answered Nov 30 '20 at 08:45

phuclv

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Also, casting to FXint with C-style cast is unsafe. That is where static_cast should have been used. Except, performing any cast is unsafe in the first place and the correct solution would be make sure that FXint is a sufficiently wide integer type. – eerorika Nov 30 '20 at 09:33
This is very good but misses that one should `#include `, and that then only guarantees to declare these types within namespace `std::`, not the global namespace. – underscore_d Nov 30 '20 at 09:48

Cast void pointer to int

1 Answers1