64 bit ntohl() in C++ ?

Question

The man pages for htonl() seem to suggest that you can only use it for up to 32 bit values. (In reality, ntohl() is defined for unsigned long, which on my platform is 32 bits. I suppose if the unsigned long were 8 bytes, it would work for 64 bit ints).

My problem is that I need to convert 64 bit integers (in my case, this is an unsigned long long) from big endian to little endian. Right now, I need to do that specific conversion. But it would be even nicer if the function (like ntohl()) would NOT convert my 64 bit value if the target platform WAS big endian. (I'd rather avoid adding my own preprocessor magic to do this).

What can I use? I would like something that is standard if it exists, but I am open to implementation suggestions. I have seen this type of conversion done in the past using unions. I suppose I could have a union with an unsigned long long and a char[8]. Then swap the bytes around accordingly. (Obviously would break on platforms that were big endian).

Thanks in advance :).

Answer 1

To detect your endian-ness, use the following union:

union {
    unsigned long long ull;
    char c[8];
} x;
x.ull = 0x0123456789abcdef; // may need special suffix for ULL.

Then you can check the contents of x.c[] to detect where each byte went.

To do the conversion, I would use that detection code once to see what endian-ness the platform is using, then write my own function to do the swaps.

You could make it dynamic so that the code will run on any platform (detect once then use a switch inside your conversion code to choose the right conversion) but, if you're only going to be using one platform, I'd just do the detection once in a separate program then code up a simple conversion routine, making sure you document that it only runs (or has been tested) on that platform.

Here's some sample code I whipped up to illustrate it. It's been tested though not in a thorough manner, but should be enough to get you started.

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

#define TYP_INIT 0
#define TYP_SMLE 1
#define TYP_BIGE 2

static unsigned long long cvt(unsigned long long src) {
    static int typ = TYP_INIT;
    unsigned char c;
    union {
        unsigned long long ull;
        unsigned char c[8];
    } x;

    if (typ == TYP_INIT) {
        x.ull = 0x01;
        typ = (x.c[7] == 0x01) ? TYP_BIGE : TYP_SMLE;
    }

 

    if (typ == TYP_SMLE)
        return src;

    x.ull = src;
    c = x.c[0]; x.c[0] = x.c[7]; x.c[7] = c;
    c = x.c[1]; x.c[1] = x.c[6]; x.c[6] = c;
    c = x.c[2]; x.c[2] = x.c[5]; x.c[5] = c;
    c = x.c[3]; x.c[3] = x.c[4]; x.c[4] = c;
    return x.ull;
}

int main (void) {
    unsigned long long ull = 1;
    ull = cvt (ull);
    printf ("%llu\n",ull);
    return 0;
}

Keep in mind that this just checks for pure big/little endian. If you have some weird variant where the bytes are stored in, for example, {5,2,3,1,0,7,6,4} order, cvt() will be a tad more complex. Such an architecture doesn't deserve to exist, but I'm not discounting the lunacy of our friends in the microprocessor industry :-)

Answer 2

some BSD systems has betoh64 which does what you need.

Answer 3

I like the union answer, pretty neat. Typically I just bit shift to convert between little and big endian, although I think the union solution has fewer assignments and may be faster:

//note UINT64_C_LITERAL is a macro that appends the correct prefix
//for the literal on that platform
inline void endianFlip(unsigned long long& Value)
{
   Value=
   ((Value &   UINT64_C_LITERAL(0x00000000000000FF)) << 56) |
   ((Value &   UINT64_C_LITERAL(0x000000000000FF00)) << 40) |
   ((Value &   UINT64_C_LITERAL(0x0000000000FF0000)) << 24) |
   ((Value &   UINT64_C_LITERAL(0x00000000FF000000)) << 8)  |
   ((Value &   UINT64_C_LITERAL(0x000000FF00000000)) >> 8)  | 
   ((Value &   UINT64_C_LITERAL(0x0000FF0000000000)) >> 24) |
   ((Value &   UINT64_C_LITERAL(0x00FF000000000000)) >> 40) |
   ((Value &   UINT64_C_LITERAL(0xFF00000000000000)) >> 56);
}

Then to detect if you even need to do your flip without macro magic, you can do a similiar thing as Pax, where when a short is assigned to 0x0001 it will be 0x0100 on the opposite endian system.

So:

unsigned long long numberToSystemEndian
(
    unsigned long long In, 
    unsigned short SourceEndian
)
{
   if (SourceEndian != 1)
   {
      //from an opposite endian system
      endianFlip(In);
   }
   return In;
}

So to use this, you'd need SourceEndian to be an indicator to communicate the endianness of the input number. This could be stored in the file (if this is a serialization problem), or communicated over the network (if it's a network serialization issue).

Answer 4

Take a look at my reply at this other question

Answer 5

An easy way would be to use ntohl on the two parts seperately:

unsigned long long htonll(unsigned long long v) {
    union { unsigned long lv[2]; unsigned long long llv; } u;
    u.lv[0] = htonl(v >> 32);
    u.lv[1] = htonl(v & 0xFFFFFFFFULL);
    return u.llv;
}

unsigned long long ntohll(unsigned long long v) {
    union { unsigned long lv[2]; unsigned long long llv; } u;
    u.llv = v;
    return ((unsigned long long)ntohl(u.lv[0]) << 32) | (unsigned long long)ntohl(u.lv[1]);
}

Answer 6

How about:

#define ntohll(x) ( ( (uint64_t)(ntohl( (uint32_t)((x << 32) >> 32) )) << 32) | 
    ntohl( ((uint32_t)(x >> 32)) ) )                                        
#define htonll(x) ntohll(x)