SPARC and x86 GCC different result of one C code

Question

One C code bring different result on “SPARC Solaris 5.9” and “Linux OpenSuse 12.1 i686 (x86)”.

#include <stdio.h>

int main(int argc, char* argv[])
{
  char Cmd = '\x00';

  char tmp[2];
  char* TempBuff = &tmp;

  *(short*)TempBuff = (Cmd << 8) | 0x5;

  printf("Out: First byte:0x%02X, Second byte: 0x%02X\n", *(TempBuff), *(TempBuff+1) );

  return 0;
}

Compiling: gcc cshort.c –o cshort

On “Linux OpenSuse 12.1 i686 (x86)”:> Out: First byte:0x05, Second byte: 0x00

On “SPARC Solaris 5.9”:> Out: First byte:0x00, Second byte: 0x05

Why, why we received different outcome?

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Details of environment:

“SPARC Solaris 5.9”:

uname –a: SunOS V245-1 5.9 Generic_118558-34 sun4u sparc SUNW,Sun-Fire-V245.

psrinfo –v: The sparcv9 processor operates at 1504 MHz, and has a sparcv9 floating point processor.

gcc version 3.4.6

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“Linux OpenSuse 12.1 i686 (x86)”:

uname –a: Linux linux-755z.site 3.1.10-1.19-desktop #1 SMP PREEMPT Mon Feb 25 10:32:50 UTC 2013 (f0b13a3) i686 i686 i386 GNU/Linux

cat /proc/cpuinfo: Intel(R) Core(TM)2 Duo CPU T8100 @ 2.10GHz

gcc version 4.6.2 (SUSE Linux)

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Below attached disassembled code of both versions.

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Answer 1

If you cast an 8 bit array type to short (16 bit) on a little endian platform, you'll get a different result than what you get on a big endian platform when doing the same.

The compiler can't help you with that, since that is just the nature of endianess...

Answer 2

Yeah, someone write abstruse and not safe code, I correct it:

//*(short*)TempBuff = (Cmd << 8) | 0x5;
TempBuff[0] = 0x5;
TempBuff[1] = Cmd;

Thank all for help.