Nanakos Chrysostomos wrote:
> i am searching for a few hours now the endianness in the x86
> environment,and i have the following
> snippets of code,which in some places i cant understand.Please help me!!!
>
> endian.c
> ---------
> #include <stdio.h>
> #include <fcntl.h>
> #include <sys/types.h>
>
>
> 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. x86 is little-endian, which means that multi-byte integers are
stored in memory with the lowest byte first, e.g. the integer
0x41424344 (1094861636 decimal) is stored 0x44,0x43,0x42,0x41.
> The same as above.Should i assume that an internal operation in printf is
> doing this???
No. It's the way that the CPU reads/writes integers to/from memory.
> I also used the above assembly example,to see what
> happens.Memory-to-memory movements (with push & pop) dont inherit the
> little-endian way.Is this
> happens only from memory-to-register and the opposite????
Yes. Endianness doesn't matter if you are just moving bytes around. It
only matters if you are treating a block of 2/4/8 bytes as a
16/32/64-bit integer.
--
Glynn Clements <[EMAIL PROTECTED]>
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