Question

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Little vs Big Endianess: How to interpret the test

Is there an easy method to test code with gcc or any online compiler like ideone for big endian? I don’t want to use qemu or virtual machines

EDIT

Can someone explain the behavior of this piece of code on a system using big endian?

#include <stdio.h>
#include <string.h>
#include <stdint.h>

int main (void)
{
    int32_t i;
    unsigned char u[4] = {'a', 'b', 'c', 'd'};

    memcpy(&i, u, sizeof(u));
    printf("%d\n", i);
    memcpy(u, &i, sizeof(i));
    for (i = 0; i < 4; i++) {
        printf("%c", u[i]);
    }
    printf("\n");
    return 0;
}

Answer 1

As a program?

#include <stdio.h>
#include <stdint.h>

int main(int argc, char** argv) {
    union {
       uint32_t word;
       uint8_t bytes[4];
    } test_struct;
    test_struct.word = 0x1;
    if (test_struct.bytes[0] != 0)
        printf("little-endian\n");
    else
        printf("big-endian\n");
    return 0;
}

On a little-endian architecture, the least significant byte is stored first. On a big-endian architecture, the most-significant byte is stored first. So by overlaying a uint32_t with a uint8_t[4], I can check to see which byte comes first. See: http://en.wikipedia.org/wiki/Big_endian

GCC in particular defines the __BYTE_ORDER__ macro as an extension. You can test against __ORDER_BIG_ENDIAN__, __ORDER_LITTLE_ENDIAN__, and __ORDER_PDP_ENDIAN__ (which I didn’t know existed!) — see http://gcc.gnu.org/onlinedocs/cpp/Common-Predefined-Macros.html

See also http://en.wikipedia.org/wiki/Big_endian

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As for running code in an endianness that doesn’t match your machine’s native endianness, then you’re going to have to compile and run it on an architecture that has that different endianness. So you are going to need to cross-compile, and run on an emulator or virtual machine.

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edit: ah, I didn’t see the first printf().

The first printf will print “1633837924”, since a big-endian machine will interpret the 'a' character as the most significant byte in the int.

The second printf will just print “abcd”, since the value of u has been copied byte-by-byte back and forth from i.