c++ - How do I convert a value from host byte order to little endian?

Question

I need to convert a short value from the host byte order to little endian. If the target was big endian, I could use the htons() function, but alas - it's not.

I guess I could do:

swap(htons(val))

But this could potentially cause the bytes to be swapped twice, rendering the result correct but giving me a performance penalty which is not alright in my case.

Answer 1

Here is an article about endianness and how to determine it from IBM:

Writing endian-independent code in C: Don't let endianness "byte" you

It includes an example of how to determine endianness at run time ( which you would only need to do once )

const int i = 1;
#define is_bigendian() ( (*(char*)&i) == 0 )

int main(void) {
    int val;
    char *ptr;
    ptr = (char*) &val;
    val = 0x12345678;
    if (is_bigendian()) {
        printf(“%X.%X.%X.%X
", u.c[0], u.c[1], u.c[2], u.c[3]);
    } else {
        printf(“%X.%X.%X.%X
", u.c[3], u.c[2], u.c[1], u.c[0]);
    }
    exit(0);
}

The page also has a section on methods for reversing byte order:

short reverseShort (short s) {
    unsigned char c1, c2;

    if (is_bigendian()) {
        return s;
    } else {
        c1 = s & 255;
        c2 = (s >> 8) & 255;

        return (c1 << 8) + c2;
    }
}

;

short reverseShort (char *c) {
    short s;
    char *p = (char *)&s;

    if (is_bigendian()) {
        p[0] = c[0];
        p[1] = c[1];
    } else {
        p[0] = c[1];
        p[1] = c[0];
    }

    return s;
}