help with displaying memory addresses

here's what i'm trying to do. I want to input a number into a variable, (int,float,double).. i want to display it's address, and display the bytes in big endian form.. (intel machines are little endian)

so, i can print the address, but &variable.. the problem is, i need the address to have spaces...

so for example, i can print "ox454567, but i want it to be 0x 45 45 67. to do this i need to go through memory byte by byte.

So, how can i go through memory byte by byte,, the code below is trying to do what i described..
but it's not working.. it keeps printing byte values inside the address

Code:

---

void IntType(void)
{
        int num0;
        char conti0;
        char *addr0;
       

        cout << endl;
        cout << endl;
        cout << "Enter an integer:";
        cin >> num0;

       
        addr0 = (char*)(&num0);
       

       

        cout << endl;
        cout << endl;
        cout << "Address:0x";
        cout << hex << addr0;

        cout << "          Big endian representation:";
        cout << endl;
        cout << endl;
        cout << endl;
        cout << endl;
        cout << endl;
        cout << endl;

        while(1)
        {
                cout << "press 'c' to continue:";
                cin >> conti0;
                cout << endl;
                if(conti0 == 'c' || conti0 == 'C') break;
        }

        cout << endl;
        cout << endl;
        cout << endl;
        cout << endl;
        cout << endl;
        cout << endl;
        cout << endl;

}

---

Seems a bit odd, but whatever. I made a little bit of code, it doesn't use std::cout but rather printf with the %x format specifier, cheating. Maybe someone would help with the std::cout implementation.

Code:

---

#include <stdio.h>
int main()
{
        int n, *p = &n, j = 0x000000ff;
        for(int i = 0; i < 32; i += 8)
                printf("%02.2\n", (unsigned int) p & (j << i));
        return 0;
}

---

This one assumes a pointer is 32 bits, like an unsigned int.

Code:

---

#include <iostream>

int main()
{
  int p;
  std::cout << std::hex << reinterpret_cast<unsigned long>(&p) << std::endl;
}

---

Of course, int* and unsigned long are not guaranteed to have the same size. Unsigned long is just likely to be a safe bet.

i'm still confised

i'm still confused

before you got confused did you at least try it.

well, i don't really understand it, I mean i can just copy and paste it. I'm not sure which variables in that thing correspond to my variables..

Code:

---

int temp;
int num; // the address of this variable is what i want to print
char *addr;
char **byt;

num = 6;
addr = #
byt = &addr;

temp  = byt;
cout << temp << " ";

byt++;

temp = byt;

cout << temp << " ";

---

i'm trying to do something like that.. not sure if that works though

okay.. i got it!!! i did something similar to what Tonto did.. it took me a while to realize he was bit shifting and masking out bits and stuff.. :D :D