Separating digits of integer using pointers - c++

Question

I have an integer(i) occupying 4 bytes and i am assuming that it is stored in the memory like this, with starting address as 1000,
If i write int*p=&i;
p now stores the starting address which is 1000 here.
if i increment p it points to the address 1004.
Is there any way to traverse the address 1000, 1001, 1002 and 1003 so that i can separate and print the digits 1 ,5,2,6 using pointers?
Please help..... :( (newbie)
My assumption of storage maybe wrong Can anyone please help me correct it? :(
EDIT 1
According to the answer given by Mohit Jain below and suggestions by others,
unsigned char *cp = reinterpret_cast<unsigned char *>(&i);
for(size_t idx = 0; idx < sizeof i; ++idx) {
cout << static_cast<int>(cp[idx]);
}
I am getting the answer as
246 5 0 0 .
I realized that the way I was assuming the memory structure was wrong,
So is there no way to get the actual digits using pointers??

Answer 1

An int with the value 1526 will not normally be stored as four bytes with the values 1, 5, 2 and 6.
Instead, it'll be stored in binary. Assuming a little-endian machine, the bytes will have the values: 0, 0, 5, 246 (and if it's big-endian, you'll get the same values in the reverse order). The reason for those numbers is that it can store values from 0 to 255 in each byte. Therefore, it's stored as 5 * 256 + 246. When dealing with values in memory like this, it's often convenient (and quite common) to use hexadecimal instead of decimal, in which case you'd be looking at it as 0x05F6.
The usual way to get decimal digits involves more math than pointers. For example, the least significant digit will be the remainder after dividing the value by 10.

Answer 2

To list the memory contents
Using pointer (endian-ness dependent output)
unsigned char *cp = reinterpret_cast<unsigned char *>(&i);
for(size_t idx = 0; idx < sizeof i; ++idx) {
cout << static_cast<int>(cp[idx]);
}
Without using pointer (endian-ness independent output), because digits are not stored the way you assume.
int copy = i;
unsigned int mask = (1U << CHAR_BIT) - 1U;
for(size_t idx = 0; idx < sizeof i; ++idx) {
cout << (copy & mask);
copy >>= CHAR_BIT;
}
To list the digits
If you want the digits of integer using pointer you should first convert the integer to a string:
std::string digits = std::to_string(i); // You can alternatively use stringstream
char *p = digits.c_str();
for(size_t idx = 0; idx < digits.length(); ++idx) cout << (*p++);

Answer 3

You can cast the pointer to (char *) and increment that pointer to point to beginning of individual bytes. However, your assumption of storage is wrong, so you will not get the digits like that.

Answer 4

As I can see you want to extract each digit of a number.
To achieve it You need to:
get reminder of i divided by 10. Do it like this: const int r = i % 10;
divide i by 10: i /= 10;
if i is not 0, go to 1.
Implementation (not tested) could be like this:
do
{
const int r = i % 10;
// do anything you need with r
i /= 10;
} while (i > 0);
This will give you each digit starting from the less significant.