C program to check little vs. big endian [duplicate]

Question

Possible Duplicate:
C Macro definition to determine big endian or little endian machine?

int main() {   int x = 1;    char *y = (char*)&x;    printf("%c\n",*y+48); } 

If it's little endian it will print 1. If it's big endian it will print 0. Is that correct? Or will setting a char* to int x always point to the least significant bit, regardless of endianness?

Answer 1

In short, yes.

Suppose we are on a 32-bit machine.

If it is little endian, the x in the memory will be something like:

       higher memory           ----->     +----+----+----+----+     |0x01|0x00|0x00|0x00|     +----+----+----+----+     A     |    &x 

so (char*)(&x) == 1, and *y+48 == '1'. (48 is the ascii code of '0')

If it is big endian, it will be:

    +----+----+----+----+     |0x00|0x00|0x00|0x01|     +----+----+----+----+     A     |    &x 

so this one will be '0'.

Answer 2

The following will do.

unsigned int x = 1; printf ("%d", (int) (((char *)&x)[0])); 

And setting &x to char * will enable you to access the individual bytes of the integer, and the ordering of bytes will depend on the endianness of the system.