How is a pointer stored in memory? - memory

Question

I am working on an OS project and I am just wondering how a pointer is stored in memory? I understand that a pointer is 4 bytes, so how is the pointer spread amongst the 4 bytes?
My issue is, I am trying to store a pointer to a 4 byte slot of memory. Lets say the pointer is 0x7FFFFFFF. What is stored at each of the 4 bytes?

Answer 1

The way that pointer is stored is same as any other multi-byte values. The 4 bytes are stored according to the endianness of the system. Let's say the address of the 4 bytes is below:
Big endian (most significant byte first):
Address Byte
0x1000 0x7F
0x1001 0xFF
0x1002 0xFF
0x1003 0xFF
Small endian (least significant byte first):
Address Byte
0x1000 0xFF
0x1001 0xFF
0x1002 0xFF
0x1003 0x7F
Btw, 4 byte address is 32-bit system. 64-bit system has 8 bytes addresses.
EDIT:
To reference each individual part of the pointer, you need to use pointer. :)
Say you have:
int i = 0;
int *pi = &i; // say pi == 0x7fffffff
int **ppi = π // from the above example, int ppi == 0x1000
Simple pointer arithmetic would get you the pointer to each byte.

Answer 2

You should read up on Endianness. Normally you wouldn't work with just one byte of a pointer at a time, though, so the order of the bytes isn't relevant.
Update: Here's an example of making a fake pointer with a known value and then printing out each of its bytes:
#include <stdio.h>
int main(int arc, char* argv[]) {
int *p = (int *) 0x12345678;
unsigned char *cp = (unsigned char *) &p;
int i;
for (i = 0; i < sizeof(p); i++)
printf("%d: %.2x\n", i, cp[i]);
return 0;
}