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#include <stdio.h>
		int main(){
		    
		   unsigned int i = 1;
		   char *c = (char*)&i;
		   if (*c)   
		       printf("Little endian\n");
		   else
		       printf("Big endian\n");
		   getchar();
		   return 0;
		}

- pradegup October 25, 2012 | Flag Reply

int a = 0xabcdef;
char *c = (char *)&a;
if (*c == a)
printf ("Big Endian");
else
printf ("Little Endian");

- Vikas June 02, 2012 | Flag Reply

Dump the bit representation for an int set as 1 in GDB. If big endian, the 1 will be on the right end, whereas if it is little endian, the 1 will be on the left one. Most computers today (x86/x86-64 architecture) are little endian.

- JYChen October 25, 2012 | Flag Reply

take an integer assign 1 to it.
take a character pointer and assign it to the address of the int check first and last value

int a = 1;
char *c = (char*)&a;
if (*c == 1)
----
else if (*(c+3) == 1)
-----

else

error

- soumyajuit May 08, 2012 | Flag Reply

how about setting the value of a character as 1 and then right shifting the variable. Endian ness can be found out by OR ing the value with 0.

- rahulbmv October 15, 2012 | Flag Reply

enum { BigEndian, LittleEndian };

int endianness(void)
{
union
{
int i;
char b[sizeof(int)];
} u;
u.i = 0x01020304;
return (u.b[0] == 0x01) ? BigEndian : LittleEndian;
}

- Anand Madhab October 25, 2012 | Flag Reply

{{{const int i=1; if (*(char*)i == 0) { //big-endian : 32 bits stored as B0B1B2B3. So you are interpreting B31 as the char. } else {//should be 1 //little endian : 32 bits stored as B31B30...B1B0. char ptr points to B0. } - bleh October 25, 2012 | Flag Reply