# Little and Big Endian Mystery

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What are these?Â
Little and big endian are two ways of storing multibyte data-types ( int, float, etc). In little endian machines, last byte of binary representation of the multibyte data-type is stored first. On the other hand, in big endian machines, first byte of binary representation of the multibyte data-type is stored first.Â
Suppose integer is stored as 4 bytes (For those who are using DOS-based compilers such as C++ 3.0, integer is 2 bytes) then a variable x with value 0x01234567 will be stored as following.
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How to see memory representation of multibyte data types on your machine?Â
Here is a sample C code that shows the byte representation of int, float and pointer.Â
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## C++

 `#include ` ` `  `/* function to show bytes in memory, from location start to start+n*/` `void` `show_mem_rep(``char` `*start, ``int` `n)` `{` `    ``int` `i;` `    ``for` `(i = 0; i < n; i++)` `         ``printf``(``" %.2x"``, start[i]);` `    ``printf``(``"\n"``);` `}` ` `  `/*Main function to call above function for 0x01234567*/` `int` `main()` `{` `   ``int` `i = 0x01234567;` `   ``show_mem_rep((``char` `*)&i, ``sizeof``(i));` `   ``getchar``();` `   ``return` `0;` `}`

## C

 `#include ` ` `  `/* function to show bytes in memory, from location start to start+n*/` `void` `show_mem_rep(``char` `*start, ``int` `n)` `{` `    ``int` `i;` `    ``for` `(i = 0; i < n; i++)` `         ``printf``(``" %.2x"``, start[i]);` `    ``printf``(``"\n"``);` `}` ` `  `/*Main function to call above function for 0x01234567*/` `int` `main()` `{` `   ``int` `i = 0x01234567;` `   ``show_mem_rep((``char` `*)&i, ``sizeof``(i));` `   ``getchar``();` `   ``return` `0;` `}`

## Java

 `import` `java.nio.ByteBuffer;`   `public` `class` `MemoryRepresentation {` `    ``public` `static` `void` `main(String[] args) {` `        ``// Create an integer value with a hexadecimal representation (0x01234567)` `        ``int` `i = ``0x01234567``;`   `        ``// Convert the integer to its byte representation, reversing the byte order` `        ``byte``[] reversedBytes = reverseArray(ByteBuffer.allocate(``4``).putInt(i).array());`   `        ``// Display the memory representation of the reversed bytes` `        ``showMemRep(reversedBytes);` `    ``}`   `    ``// Display the memory representation of a byte array` `    ``public` `static` `void` `showMemRep(``byte``[] data) {` `        ``for` `(``byte` `b : data) {` `            ``// Print each byte as a two-digit hexadecimal number` `            ``System.out.printf(``"%02x "``, b & ``0xFF``);` `        ``}` `        ``System.out.println();` `    ``}`   `    ``// Reverse the order of bytes in a byte array` `    ``public` `static` `byte``[] reverseArray(``byte``[] array) {` `        ``byte``[] reversed = ``new` `byte``[array.length];` `        ``for` `(``int` `i = ``0``; i < array.length; i++) {` `            ``// Copy bytes in reverse order to the 'reversed' array` `            ``reversed[i] = array[array.length - ``1` `- i];` `        ``}` `        ``return` `reversed;` `    ``}` `}`

## Python3

 `import` `ctypes`   `# function to show bytes in memory, from location start to start+n`     `def` `show_mem_rep(start, n):` `    ``# create a ctypes array of unsigned bytes from the buffer` `    ``arr ``=` `(ctypes.c_ubyte ``*` `n).from_address(ctypes.addressof(start))` `    ``for` `i ``in` `range``(n):` `        ``# use hex() to convert integer to hexadecimal string and slice to remove the "0x" prefix` `        ``# use rjust() to right-justify the string with 2 spaces` `        ``print``(``hex``(arr[i])[``2``:].rjust(``2``, ``"0"``), end``=``" "``)` `    ``print``()`     `# Main function to call above function for 0x01234567` `if` `__name__ ``=``=` `"__main__"``:` `    ``# create a 4-byte integer with value 0x01234567` `    ``i ``=` `0x01234567` `    ``# convert integer i to a byte array in little-endian byte order` `    ``i_bytes_le ``=` `i.to_bytes(ctypes.sizeof(ctypes.c_int), byteorder``=``"little"``)` `    ``# convert integer i to a byte array in big-endian byte order` `    ``i_bytes_be ``=` `i.to_bytes(ctypes.sizeof(ctypes.c_int), byteorder``=``"big"``)` `    ``# create a writable buffer using ctypes.create_string_buffer() and copy the bytes into it` `    ``buf_le ``=` `ctypes.create_string_buffer(i_bytes_le)` `    ``buf_be ``=` `ctypes.create_string_buffer(i_bytes_be)` `    ``# pass the buffer to show_mem_rep` `    ``print``(``"Little Endian: "``, end``=``"")` `    ``show_mem_rep(buf_le, ctypes.sizeof(ctypes.c_int))` `    ``print``(``"Big Endian: "``, end``=``"")` `    ``show_mem_rep(buf_be, ctypes.sizeof(ctypes.c_int))`

## C#

 `using` `System;`   `class` `Program` `{` `    ``static` `void` `Main(``string``[] args)` `    ``{` `        ``int` `i = 0x01234567;` `        ``ShowMemRep(BitConverter.GetBytes(i));` `        ``Console.ReadLine();` `    ``}`   `    ``static` `void` `ShowMemRep(``byte``[] bytes)` `    ``{` `        ``for` `(``int` `i = 0; i < bytes.Length; i++)` `        ``{` `            ``Console.Write(``" {0:X2}"``, bytes[i]);` `        ``}` `        ``Console.WriteLine();` `    ``}` `}`

## Javascript

 `// Function to show bytes in memory, from location start to start+n` `function` `show_mem_rep(start, n) {` `    ``for` `(let i = 0; i < n; i++) {` `        ``// Print the byte at index i in hexadecimal format with 2 digits` `        ``console.log(` \${start[i].toString(16).padStart(2, ``'0'``)}`);` `    ``}` `    ``console.log(``"\n"``);` `}`   `// Main function to call above function for 0x01234567` `function` `main() {` `    ``// In JavaScript, we don't have specific data types like int or char, so we use numbers` `    ``let i = 0x01234567;`   `    ``// When accessing memory in JavaScript, we use Typed Arrays` `    ``// Here, we create a Uint8Array view of the memory where `i` is stored` `    ``let byteArray = ``new` `Uint8Array(``new` `Uint32Array([i]).buffer);`   `    ``// Call the show_mem_rep function with the byteArray and its length` `    ``show_mem_rep(byteArray, byteArray.length);` `}`   `// Call the main function` `main();`

Output

``` 67 45 23 01

```

Time Complexity: O(1)

Auxiliary Space: O(1)

When above program is run on little endian machine, gives “67 45 23 01” as output, while if it is run on big endian machine, gives “01 23 45 67” as output.
Is there a quick way to determine endianness of your machine?Â
There are n no. of ways for determining endianness of your machine. Here is one quick way of doing the same.Â
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## C++

 `#include ` `using` `namespace` `std;` `int` `main() ` `{ ` `    ``unsigned ``int` `i = 1; ` `    ``char` `*c = (``char``*)&i; ` `    ``if` `(*c) ` `        ``cout<<``"Little endian"``; ` `    ``else` `        ``cout<<``"Big endian"``; ` `    ``return` `0; ` `} `   `// This code is contributed by rathbhupendra`

## C

 `#include ` `int` `main() ` `{` `   ``unsigned ``int` `i = 1;` `   ``char` `*c = (``char``*)&i;` `   ``if` `(*c)    ` `       ``printf``(``"Little endian"``);` `   ``else` `       ``printf``(``"Big endian"``);` `   ``getchar``();` `   ``return` `0;` `}`

## Java

 `public` `class` `EndiannessCheck {` `    ``public` `static` `void` `main(String[] args) {` `        ``int` `i = ``1``; ``// Initialize an integer with value 1` `        ``byte``[] byteArray = toByteArray(i); ``// Convert integer to byte array`   `        ``// Check endianness` `        ``if` `(byteArray[``0``] == ``1``) {` `            ``System.out.println(``"Little endian"``);` `        ``} ``else` `{` `            ``System.out.println(``"Big endian"``);` `        ``}` `    ``}`   `    ``// Utility method to convert an integer to a byte array` `    ``private` `static` `byte``[] toByteArray(``int` `value) {` `        ``byte``[] byteArray = ``new` `byte``[``4``];` `        ``for` `(``int` `i = ``0``; i < ``4``; i++) {` `            ``byteArray[i] = (``byte``) (value >>> (i * ``8``));` `        ``}` `        ``return` `byteArray;` `    ``}` `}`

## C#

 `using` `System;`   `class` `Program` `{` `    ``static` `void` `Main(``string``[] args)` `    ``{` `        ``uint` `i = 1;` `        ``byte``[] bytes = BitConverter.GetBytes(i); ``// Convert the integer to bytes`   `        ``if` `(bytes[0] != 0)` `        ``{` `            ``Console.WriteLine(``"Little endian"``);` `        ``}` `        ``else` `        ``{` `            ``Console.WriteLine(``"Big endian"``);` `        ``}` `    ``}` `}`

Output

```Little endian

```

Output:Â
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`Little endian`

Time Complexity: O(1)

Auxiliary Space: O(1)

In the above program, a character pointer c is pointing to an integer i. Since size of character is 1 byte when the character pointer is de-referenced it will contain only first byte of integer. If machine is little endian then *c will be 1 (because last byte is stored first) and if the machine is big endian then *c will be 0.Â
Does endianness matter for programmers?Â
Most of the times compiler takes care of endianness, however, endianness becomes an issue in following cases.
It matters in network programming: Suppose you write integers to file on a little endian machine and you transfer this file to a big-endian machine. Unless there is little endian to big endian transformation, big endian machine will read the file in reverse order. You can find such a practical example here.
Standard byte order for networks is big endian, also known as network byte order. Before transferring data on network, data is first converted to network byte order (big endian).Â
Sometimes it matters when you are using type casting, below program is an example.
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## C++

 `#include ` `using` `namespace` `std;`   `int` `main() {` `unsigned ``char` `arr[2] = {0x01, 0x00};` `unsigned ``short` `int` `x = *(unsigned ``short` `int` `*) arr;` `cout << x << endl;` `cin.get();` `return` `0;` `}`

## c

 `#include ` `int` `main()` `{` `    ``unsigned ``char` `arr[2] = {0x01, 0x00};` `    ``unsigned ``short` `int` `x = *(unsigned ``short` `int` `*) arr;` `    ``printf``(``"%d"``, x);` `    ``getchar``();` `    ``return` `0;` `}`

## Java

 `public` `class` `Main {` `    ``public` `static` `void` `main(String[] args) {` `        ``byte``[] arr = {(``byte``) ``0x01``, (``byte``) ``0x00``};` `        ``short` `x = (``short``)((arr[``1``] << ``8``) | (arr[``0``] & ``0xFF``));` `        ``System.out.println(x);` `    ``}` `}` `// This code is contributed by shivhack999`

## C#

 `using` `System;`   `class` `Program` `{` `    ``static` `void` `Main()` `    ``{` `        ``// Create a byte array with the same values as the C++ array` `        ``byte``[] arr = { 0x01, 0x00 };`   `        ``// Interpret the byte array as an unsigned short integer` `        ``ushort` `x = BitConverter.ToUInt16(arr, 0);`   `        ``// Output the result` `        ``Console.WriteLine(x);`   `        ``// Wait for user input before exiting (simulating cin.get())` `        ``Console.ReadLine();` `    ``}` `}`

Output

```1

```

Time Complexity: O(1)

Auxiliary Space: O(1)

In the above program, a char array is typecasted to an unsigned short integer type. When I run above program on little endian machine, I get 1 as output, while if I run it on a big endian machine I get 256. To make programs endianness independent, above programming style should be avoided.Â
What are bi-endians?Â
Bi-endian processors can run in both modes little and big endian.
What are the examples of little, big endian and bi-endian machines ?Â
Intel based processors are little endians. ARM processors were little endians. Current generation ARM processors are bi-endian.
Motorola 68K processors are big endians. PowerPC (by Motorola) and SPARK (by Sun) processors were big endian. Current version of these processors are bi-endians.Â
Does endianness affects file formats?Â
File formats which have 1 byte as a basic unit are independent of endianness e.g., ASCII files. Other file formats use some fixed endianness format e.g, JPEG files are stored in big endian format.Â
Which one is better — little endian or big endian?Â
The term little and big endian came from Gulliverâ€™s Travels by Jonathan Swift. Two groups could not agree by which end an egg should be opened -a- the little or the big. Just like the egg issue, there is no technological reason to choose one-byte ordering convention over the other, hence the arguments degenerate into bickering about sociopolitical issues. As long as one of the conventions is selected and adhered to consistently, the choice is arbitrary.
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