How do I cast a pointer to an int

Question

I\'m trying to store the value of an address in a non pointer int variable, when I try to convert it I get the compile error \"invalid conversion from \'int*\' to \'int\'\" this

Answer 1

Why are you trying to do that, anyway you just need to cast, for C code :

thatvalue = (int)ip;

If your writing C++ code, it is better to use reinterpret_cast

Answer 2

I'd suggest using reinterpret_cast:

thatvalue = reinterpret_cast<intptr_t>(ip);

Answer 3

You can do this:

int a_variable = 0;

int* ptr = &a_variable;

size_t ptrValue = reinterpret_cast<size_t>(ptr);

Answer 4

I was able to use the C union statement to achieve what you were looking for. It will of course be compiler dependent, but it worked for me like you would presume it should (Linux, g++).

union {
    int i;
    void *p;
} mix;

mix.p = ip;
cout << mix.i << endl;

On my particular instance my int is 32 bit and the pointer is 48 bit. When assigning the pointer, the integer value i will represent the lowest 32 bit of the pointer.

Answer 5

int may not be large enough to store a pointer.

You should be using intptr_t. This is an integer type that is explicitly large enough to hold any pointer.

    intptr_t thatvalue = 1;

    // stuff

    thatvalue = reinterpret_cast<intptr_t>(ip);
                // Convert it as a bit pattern.
                // It is valid and converting it back to a pointer is also OK
                // But if you modify it all bets are off (you need to be very careful).