Displaying a pointer in hex

Probably the easiest way is to use a stream and output the pointer value casted to void*, like this:

#include <iostream>
using namespace std;
int main()
{
    char sz[] = "some chars";
    cout << static_cast<void *>(sz) << endl;
    return 0;
}

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For dialog, you may want to use <sstream> instead:

#include <iostream>
#include <sstream>

using namespace std;

int main()
{
    char sz[] = "some chars";

    ostringstream oss;
    oss << static_cast<void *>(sz);

    cout << oss.str() << endl;
    return 0;
}

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Here displayed also via cout; however, oss.str() is the string representation of the pointer value. For Windows dialog, you probly need the C-string. Then use oss.str().c_str().

For displaying integer values in hex, include the standard header iomanip that gives you the hex manipulator, setfill('0'), and setw(8). Using them for any stream, the next integer will be displayed in hex representation using 8 characters, left padded by zeros:

#include <iomanip>
    ...
    ostringstream oss;
    oss << << hex << setfill('0') << setw(8) << theValue;

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