Endian conversion in C signed

Question

I was trying to write signed integer value in a file in big endian.

I'm using this code :

int32_t swap_int32( int32_t val )
{
    val = ((val << 8) & 0xFF00FF00) | ((val >> 8) & 0xFF00FF ); 
    return (val << 16) | ((val >> 16) & 0xFFFF);
}

It works fine with negative value but I've some issues with positive value :

-5 gives me :

$> hexdump OUTPUT
ff ff ff fb

Which is correct

but 1337 gives me :

  $> hexdump OUTPUT
     00 00 05 39

Instead of (expected) 00 00 39 05.

Is there anyway to treat both cases and get correct result ? Thanks you.

It has nothing to do with positive and negative, but rather absolute values greater than 255. — Mark Ransom, Dec 16 '11 at 17:12
1337 is 0x539 so in big endian it's 00 00 05 39, not 00 00 39 05. — Artefacto, Dec 16 '11 at 17:17
tutorial here: http://www.gamedev.net/page/resources/_/technical/game-programming/writing-endian-independent-code-in-c-r2091 — Hybrid, Dec 16 '11 at 17:18
don't do shifts with signed values unless you know exactly what you are doing. — Jens Gustedt, Dec 16 '11 at 17:23
possible duplicate of [Signed Integer Network and Host Conversion](http://stackoverflow.com/questions/4878781/signed-integer-network-and-host-conversion) — Jens Gustedt, Dec 16 '11 at 17:25

Paul Rubel · Answer 1 · 2011-12-16T17:23:57.823

Is there some problem with using htonl?

   #include <arpa/inet.h>

   uint32_t htonl(uint32_t hostlong);

The values in the bytes should stay the same, so even though you want a signed value and htonl take an unsigned as long as you read it back into a signed type you shouldn't lose data. (As noted by mctylr, network order is big-endian.)

#include <arpa/inet.h>
#include <stdio.h>

int main(int argc, char** argv) {
  int32_t pos = 1337;
  int32_t neg = -123;
  printf("host-endian:  %d => 0x%08x  %d => 0x%08x \n", 
         pos, pos, neg, neg);

  uint32_t big_endian_pos = htonl(pos);
  uint32_t big_endian_neg = htonl(neg);

  printf("Big-endian:   %d => 0x%08x %d => 0x%08x \n", 
         pos, big_endian_pos, neg, big_endian_neg);

}

And the output on my little-endian host:

[tmp]$ gcc endian.c  && ./a.out 
host-endian:  1337 => 0x00000539  -123 => 0xffffff85 
Big-endian:   1337 => 0x39050000  -123 => 0x85ffffff

To clarify, *network byte order* is **big** endian. So on either an little endian or big endian host (system), the network long (unsigned 32-bit integer) is written in big endian byte ordering. — mctylr, Dec 16 '11 at 17:19

score 1 · Answer 2 · answered Dec 16 '11 at 17:17

1

At a minimum, you should cast to "unsigned" inside your function.

As an alternative, you can always use the network macros "htons()" (16-bit) and "htonl()" (32-bit) to convert host to network (big-endian) byte order.

answered Dec 16 '11 at 17:17

paulsm4

114,292
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Good point. `x >> y` might be an arithmetic shift when `x` is signed, depending on the compiler: http://stackoverflow.com/questions/7622/shift-operator-in-c – bk1e Dec 16 '11 at 18:00

score 0 · Answer 3 · answered Dec 16 '11 at 17:19

0

Network byte order is big endian.

POSOX UNIX has htonl(), htons() and friends to do just what you require. They guarantee big endian from any system.

answered Dec 16 '11 at 17:19

jim mcnamara

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score 0 · Answer 4 · edited May 23 '17 at 12:27

On *BSD and Linux systems you can use byteorder(9) functions including bswap_32() when used on a little endian host.

#include <sys/endian.h>  /* other platforms may need: sys/param.h endian.h or byteswap.h */
#include <stdint.h>

uint32_t in, out;

out = bswap_32(in);

On little endian MS Windows systems, _byteswap_ulong is included in stdlib.h, even though it is not standard.

Otherwise various bit twiddling examples can be easily found across the Internet.

And StackOverflow

convert big endian to little endian in C {without using provided func} 2182002
How do I convert between big-endian and little-endian values in C++? 105252
Fast little-endian to big-endian conversion in ASM 1358747 - 2 out of 3 answers are not C# oriented

Endian conversion in C signed

4 Answers4