Reputation: 834
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?
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
“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)
Below attached disassembled code of both versions.
Upvotes: 3
Views: 882
Reputation: 834
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.
Upvotes: -1
Reputation: 3335
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...
Upvotes: 5