Converting a pointer into an integer

'size_t' and 'ptrdiff_t' are required to match your architecture (whatever it is). Therefore, I think rather than using 'int', you should be able to use 'size_t', which on a 64 bit system should be a 64 bit type.

This discussion unsigned int vs size_t goes into a bit more detail.

Use intptr_t and uintptr_t.

To ensure it is defined in a portable way, you can use code like this:

#if defined(__BORLANDC__)
    typedef unsigned char uint8_t;
    typedef __int64 int64_t;
    typedef unsigned long uintptr_t;
#elif defined(_MSC_VER)
    typedef unsigned char uint8_t;
    typedef __int64 int64_t;
#else
    #include <stdint.h>
#endif

Just place that in some .h file and include wherever you need it.

Alternatively, you can download Microsoft’s version of the stdint.h file from here or use a portable one from here.

I'd say this is the modern C++ way:

#include <cstdint>
void *p;
auto i = reinterpret_cast<std::uintptr_t>(p);

EDIT:

The correct type to the the Integer

So the right way to store a pointer as an integer is to use the uintptr_t or intptr_t types. (See also in cppreference integer types for C99).

These types are defined in <stdint.h> for C99 and in the namespace std for C++11 in <cstdint> (see integer types for C++).

C++11 (and onwards) Version

#include <cstdint>
std::uintptr_t i;

C++03 Version

extern "C" {
#include <stdint.h>
}

uintptr_t i;

C99 Version

#include <stdint.h>
uintptr_t i;

The correct casting operator

In C there is only one cast and using the C cast in C++ is frowned upon (so don't use it in C++). In C++ there are different types of casts, but reinterpret_cast is the correct cast for this conversion (see also here).

C++11 Version

auto i = reinterpret_cast<std::uintptr_t>(p);

C++03 Version

uintptr_t i = reinterpret_cast<uintptr_t>(p);

C Version

uintptr_t i = (uintptr_t)p; // C Version

Related Questions

• What is uintptr_t data type