Make the pointer look like its pointing to desired address

[bubblesbubby]

hi,
First let me post a code here to make my question clear.

#include <iostream>

using namespace std;

int main()
{
  unsigned int *ptr;
  int val = 150;
  ptr = &val;
  cout << "pointer location is: " << ptr <<endl;
}

The code above displays a hex address pointed by the pointer, "otr". But for my code i wanted it to look like it is pointing from location 0 to 150, without changing the actual value of the pointer "ptr". Is there a way that i can achieve this?

[Daleth]

ptr = &val;
for(unsigned int I = 0; I <= ptr; I++)
   std::cout << I << std::endl;

?

BTW, ptr does not hold a value of 150. That is the value of val. ptr's value is whatever memory address val is at, and *ptr holds the value 150.

[giblit]

It points to the memory address AFAIK but you can change the pointed address's value by something like this

short value = 10 , *ptr = &value;
*ptr++;
std::cout << value << std::endl;

[bubblesbubby]

ok... Thanks....