Visualizing Little Endian using C++

I have been doing computer programming for around 8 years now and some of the topics I always brushed away included bit-level and low-level under-the-hood concepts that are often abstracted from the programmer by high level languages such as Java and C#. Over time I realized that I did not fully understand what really happens on the "metal" when it came to topics like bit manipulation.

Learning C and C++ made me realize the need to understand what happens at the low level. To help me remember what I learned and hopefully clarify to other readers, I am going to share what I learned about endianness on x86-based systems, which are Little Endian.

Jumping straight into the code, I have written a very simple to understand program that will show how bits and bytes are stored in memory on an x86 system. It is very important to remember that endianness is about bytes not bits - very important point to remember. The results displayed by the program will support that. The code (along with comments should be self explanatory). I wrote the code with simplicity in mind so that even a beginner in C++ should find it easy to read. However, if anything is not clear, feel free to comment and I will reply as soon as I can.

#include <iostream>

using namespace std;

int main() {
    // Assign a value to a 32-bit so that we can view each of the four bytes
    // and read their addresses in memory.
    unsigned int i = 0x12345678;

    // char is 1 byte, so we will use it to read each of the 4 bytes in the
    // integer memory location.
    unsigned char* c = (unsigned char*)&i;

    // Show the value of the integer and the address in memory.
    // Take note of the address for comparing with other addresses displayed
    // for each of the bytes displayed below.
    cout << endl << "Value of i :\t0x" << hex << i << endl;
    cout << "Address of i :\t" << &i << endl << endl;

    // Show the address and value of the first byte.
    cout << "Address of c : \t0x" << (int)(c) << endl;
    cout << "Value of c[0] :\t0x" << hex << (int)(*c)
         << ", Address = 0x" << hex << (int)c << endl;

    c++; // Move to the 2nd byte
    cout << "Value of c[1] :\t0x" << hex << (int)(*c)
         << ", Address = 0x" << hex << (int)c << endl;

    c++; // Move to 3rd byte.
    cout << "Value of c[2] :\t0x" << hex << (int)(*c)
         << ", Address = 0x" << hex << (int)c << endl;

    c++; // Move 4th (last) byte.
    cout << "Value of c[3] :\t0x" << hex << (int)(*c)
         << ", Address = 0x" << hex << (int)c << endl;

    return 0;
}

Running the program on my computer will show the following results:

Value of i :    0x12345678
Address of i :  0x6afee8

Address of c :  0x6afee8
Value of c[0] : 0x78, Address = 0x6afee8
Value of c[1] : 0x56, Address = 0x6afee9
Value of c[2] : 0x34, Address = 0x6afeea
Value of c[3] : 0x12, Address = 0x6afeeb

Analysis

What is important to take from the results is the following:

1. The least significant byte (LSB) of the integer is stored in a low memory address. This is what is expected since the system is little endian. The LSB is the one with value 0x78.
2. In each of the bytes, the bit order is not changed i.e., the high byte has the value 0x12 not 0x21

I have included a diagram below to show a visual representation if the bytes in memory, with low addresses at the bottom:

 <<value>>   <<adddress>>
+          +
+          +
+----------+ 
+    12    + 0x6afeeb
+----------+ 
+    34    + 0x6afeea
+----------+ 
+    56    + 0x6afee9
+----------+ 
+    78    + 0x6afee8
+----------+ 
+          +
+          +
 

That's it for this post. :)