Converting a pointer into an integer

Question

I am trying to adapt an existing code to a 64 bit machine. The main problem is that in one function, the previous coder uses a void* argument that is converted into suitable type in the function itself. A short example:

void function(MESSAGE_ID id, void* param) {     if(id == FOO) {         int real_param = (int)param;         // ...     } } 

Of course, on a 64 bit machine, I get the error:

error: cast from 'void*' to 'int' loses precision 

I would like to correct this so that it still works on a 32 bit machine and as cleanly as possible. Any idea ?

Answer 1

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 

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