SPARC and x86 GCC are different results of a C code
A piece of C code brings mixed results 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;
}
Compile: gcc cshort.c –o cshort
On “Linux OpenSuse 12.1 i686 (x86)”: > output: first byte: 0x05, second byte: 0x00
On “SPARC Solaris 5.9”: > output: first byte: 0x00, second byte: 0x05
Why, why do we receive different results?
Environmental details:
“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 runs at 1504 MHz and has a sparcv9 floating-point processor.
GCC version 3.4.6
“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).
Two versions of the disassembly code are attached below.
Solution
If you convert an 8-bit array type to short (16-bit) on a little-endian platform, you will get different results than if you did the same on a big-endian platform.
The compiler can’t help you because that’s just the nature of byte order….