On Fri, 8 Jul 2005 22:05:17 +0300 (EEST)
"Nanakos Chrysostomos" <[EMAIL PROTECTED]> wrote:
> int main()
> {
> char *filename= "endian.txt";
> unsigned long buf;
> char *k=(char *)&buf;
> int fd;
>
> fd = open("makis",O_RDONLY);
>
> read(fd,&buf,4);
>
> printf("%.4s\n",&buf);
> printf("%p\n",buf);
> printf("&buf: %p %#x %p\n",&buf,*k,k);
> return 0;
> }
>
> endian.txt
> ----------
> DBCA
>
>
> #./read
> DCBA
> 0x41424344
> &buf: 0xbffff8b0 0x44 0xbffff8b0
> #
>
>
> In the first printf everything is fine.In the second printf we see
> that the 0x44,0x43,0x42,0x41 byte-data is printed in the revserse
> order,while we can see
> that in memory it is in the right order after the read system call.Why
> this happens?Is it being internal in printf???
No, it isn't printf... it's just that x86 CPU are little endian.
When you do:
printf("%.4s\n",&buf);
you are printing these four bytes in memory-order.
When you do:
printf("%p\n",buf);
you are treating "buf" as a single number. Since your CPU is little
endian it expects the low order bytes to come first.
Another example:
#include <stdio.h>
int main(void) {
unsigned int buf = 0x11223344;
unsigned char *x = (unsigned char*)&buf;
printf("0x%hhx\n", x[0]);
return 0;
}
printf will print 0x44... simply because the 0x11223344 numer is stored
in memory in this way:
address value
&buf 0x44
&buf+1 0x33
&buf+2 0x22
&buf+3 0x11
For more info about x86 CPU look here:
http://developer.intel.com/design/pentium4/manuals/index_new.htm
--
Paolo Ornati
Linux 2.6.12.2 on x86_64
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