When I do the following, I get this error:
../src/Sample.cpp:19: error: cast from \u2018UINT8*\u2019 to \u2018UINT8\u2019 loses precision
#include <iostream>
using namespace std;
typedef unsigned char UINT8;
typedef unsigned int UINT32;
#define UNUSED(X) X=X
int main() {
UINT8 * a = new UINT8[34];
UINT32 b = reinterpret_cast<UINT8>(a);
UNUSED(b);
return 0;
}
How would I go about solving this. Keep in mind I am not trying to convert string to unsigned long, rather converting char* (the ADDRESS value ) to int.
Thanks
Solution:
Turns out that this problem has to do with the pointer size. On a 32 bit machine, the pointer size is 32 bit, and for 64 bit machine is of course 64. The above wont work on a 64 bit machine, but will on a 32 bit machine. This will work on a 64 bit machine.
#include <iostream>
#include <stdint.h>
using namespace std;
typedef uint8_t UINT8;
typedef int64_t UINT32;
#define UNUSED(X) X=X
int main() {
UINT8 * a = new UINT8[34];
UINT32 b = reinterpret_cast<UINT32>(a);
UNUSED(b);
return 0;
}
Best Answer
Assuming that sizeof(int) == sizeof(void*) you can use this code to convert:
or variations thereof. I think static_cast will work too.
Of course a has to be l-value (able to be assigned to) to get the address of it. For non-l-values, you need to use a function and the good old union trick will work: