How do I use an int string with a messagebox?

Question

I'm trying to get a messagebox to display the address of a variable by casting an int to a const CHAR*, my current dysfunctional attempt looks like this:

#include <cstdlib>
#include <iostream>
#include <windows.h>

int main()
{
 int *ip;
 int pointervalue = 1337;
 int thatvalue = 1;
 ip = &pointervalue;
 thatvalue = (int)ip;
 std::cout<<thatvalue<<std::endl;
 MessageBox (NULL, (const char*)thatvalue, NULL, NULL);
 return 0;
}

the dos box prints 2293616, the messagebox prints "9|"

Answer 1

If you're using C++11, you can also use to_string() :

MessageBox (NULL, std::to_string(thatvalue).c_str(), NULL, NULL);

Your current problem is that you're just casting thatvalue to const char* , or in other words, taking the int value and converting it to a pointer , not a string (C-style or otherwise). You're getting junk printed in your message box because the const char* pointer is pointing to invalid, junk memory, and it's an unfortunate miracle it's not crashing.

Answer 2

try to use stringstream instead (include sstream)

int *ip;
int pointervalue = 1337;
int thatvalue = 1;
ip = &pointervalue;    
stringstream ss;
ss << hex << ip;
MessageBox (NULL, ss.str().c_str(), NULL, NULL);

Answer 3

Simple Casting won't do this Job.

Have a Look at the itoa function: http://www.cplusplus.com/reference/cstdlib/itoa/

/* itoa example */
#include <stdio.h>
#include <stdlib.h>

int main ()
{
  int i;
  char buffer [33];
  printf ("Enter a number: ");
  scanf ("%d",&i);
  itoa (i,buffer,10);
  printf ("decimal: %s\n",buffer);
  itoa (i,buffer,16);
  printf ("hexadecimal: %s\n",buffer);
  itoa (i,buffer,2);
  printf ("binary: %s\n",buffer);
  return 0;
}

Answer 4

Casting to const char * does not work because then it tries to interpret the int as a pointer.

If you want to avoid streams you can use snprintf like so

char buffer[20];
snprintf(buffer,20,"%d",thatValue);
MessageBox (NULL, (const char*)buffer, NULL, NULL);