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.
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.
Is there a quick way to determine endianness of your machine?
There are n no. of ways for determining endianness of your machine. Here is one quick way of doing the same.
#include <stdio.h>
int main()
{
unsigned int i = 1;
char *c = (char*)&i;
if (*c)
printf("Little endian");
else
printf("Big endian");
getchar();
return 0;
}Does endianness matter for programmers?
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 andian 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).
Sometimes it matters when you are using type casting, below program is an example.
#include <stdio.h>
int main()
{
unsigned char arr[2] = {0x01, 0x00};
unsigned short int x = *(unsigned short int *) arr;
printf("%d", x);
getchar();
return 0;
}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.
