Is there a one line macro definition to determine the endianness of the machine? I am using the following code but converting it to macro would be too long:
unsigned char test_endian( void )
{
int test_var = 1;
unsigned char *test_endian = (unsigned char*)&test_var;
return (test_endian[0] == 0);
}
If it is little-endian, it would be stored as “01 00 00 00”. The program checks the first byte by dereferencing the cptr pointer. If it equals to 0, it means the processor is big-endian(“00 00 00 01”), If it equals to 1, it means the processor is little-endian (“01 00 00 00”).
See this comment in the GCC mailing list: Bit-fields are always assigned to the first available bit, possibly constrained by other factors, such as alignment. That means that they start at the low order bit for little-endian, and the high order bit for big-endian.
Little Endian Byte Order: The least significant byte (the "little end") of the data is placed at the byte with the lowest address. The rest of the data is placed in order in the next three bytes in memory.
Explanation: Endianness is the sequential order used to interpret a range of bytes in the memory of a system.
Code supporting arbitrary byte orders, ready to be put into a file called order32.h
:
#ifndef ORDER32_H
#define ORDER32_H
#include <limits.h>
#include <stdint.h>
#if CHAR_BIT != 8
#error "unsupported char size"
#endif
enum
{
O32_LITTLE_ENDIAN = 0x03020100ul,
O32_BIG_ENDIAN = 0x00010203ul,
O32_PDP_ENDIAN = 0x01000302ul, /* DEC PDP-11 (aka ENDIAN_LITTLE_WORD) */
O32_HONEYWELL_ENDIAN = 0x02030001ul /* Honeywell 316 (aka ENDIAN_BIG_WORD) */
};
static const union { unsigned char bytes[4]; uint32_t value; } o32_host_order =
{ { 0, 1, 2, 3 } };
#define O32_HOST_ORDER (o32_host_order.value)
#endif
You would check for little endian systems via
O32_HOST_ORDER == O32_LITTLE_ENDIAN
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