How Can I Safely Convert a Pointer to an Integer on 64-Bit Machines?

Converting a Pointer into an Integer: Revisited for 64-Bit Machines

In C/C , a potential pitfall when dealing with 64-bit machines arises when handling pointers. Traditional conversion techniques, such as the one shown below, may result in precision errors:

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

Solution with Updated Casting Techniques

To rectify this issue and ensure seamless compatibility on both 32-bit and 64-bit systems, a modern C approach is recommended:

#include <cstdint>

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

Choosing the Correct Data Type

When storing a pointer as an integer, it is crucial to select the appropriate data type. For this purpose, the uintptr_t type is most suitable.

// C++11
std::uintptr_t i;

// C++03
extern "C" {
#include <stdint.h>
}
uintptr_t i;

// C99
#include <stdint.h>
uintptr_t i;

Employing the Correct Casting Operator

In C , various types of casts exist. For this specific scenario, reinterpret_cast is the most appropriate choice.

// C++11
auto i = reinterpret_cast<std::uintptr_t>(p);

// C++03
uintptr_t i = reinterpret_cast<uintptr_t>(p);

// C
uintptr_t i = (uintptr_t)p;

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