Subject: Re: B19.1: memcpy bug (NT)
From: [email-masked] ($Bill Luebkert)

23 Aug 1998 02:32:47 -0700
Message-ID: <[email-masked]>
References: <[email-masked]>
Mime-Version: 1.0
Content-Type: text/plain; charset=us-ascii
Content-Transfer-Encoding: 7bit
To: "Vincent S. Perun" <[email-masked]>
Cc: "[email-masked]" <[email-masked]>

Vincent S. Perun wrote:
> 
> Hello.  I've discovered a bug with memcpy under beta 19.1 and NT 4.0.  It
> seems the contents of a char array don't copy into an int in the right order:
> The bytes are reversed!  This test expects the output to be "ABCD" but it
> puts out "CDAB".  A char to char memcpy works just fine!
> 
> Vincent Perun
> 
> Test program:
> 
> main()
> {
>    unsigned char s1[2] = {0xAB, 0xCD};
>    unsigned short n;
>    unsigned char s2[2];
> 
>    memcpy (&n, &s1, 2);
>    printf ("n: %4.4X\n", n);
>    memcpy (&s2, &s1, 2);
>    printf ("s2: %2.2X%2.2X\n", s2[0], s2[1]);
> }

You are running this on an Intel platform, right?  This is expected 
behavior if so.

Intel uses little-endian byte storage which means that the value 
stored in a short are byte reversed from what you would have on a 
big-endian machine.  IE: bytes are stored from the right of a register 
to the first location in memory and successive bytes to the left in 
the register are stored in higher numbered memory locations.

Add this after the first printf to print it out forward:

	(void)printf ("n: %2.2X%2.2X\n", n & 0xff, (n & 0xff00) >> 8);

You may need to do a little study on machine architecture.  :)

PS: I worked at JPL for 21 years.  How's things going over there?
-- 
  ,-/-  __      _  _         $Bill Luebkert   ICQ=14439852
 (_/   /  )    // //       DBE Collectibles
  / ) /--<  o // //      Mailto:[email-masked]
-/-' /___/_<_</_</_    http://www.wgn.net/~dbe/
-
For help on using this list (especially unsubscribing), send a message to
"[email-masked]" with one line of text: "help".