How to interpret *( (char*)&a )

Question

I see a way to know the endianness of the platform is this program but I don't understand it

#include <stdio.h>
int main(void)
{
  int a = 1;
  if( *( (char*)&a ) == 1) printf("Little Endian\n");
  else printf("Big Endian\n");
  system("PAUSE");  
  return 0;
}

What does the test do?

Answer 1

An int is almost always larger than a byte and often tracks the word size of the architecture. For example, a 32-bit architecture will likely have 32-bit ints. So given typical 32 bit ints, the layout of the 4 bytes might be:

   00000000 00000000 00000000 00000001

or with the least significant byte first:

   00000001 00000000 00000000 00000000

A char* is one byte, so if we cast this address to a char* we'll get the first byte above, either

   00000000

or

   00000001

So by examining the first byte, we can determine the endianness of the architecture.