From mboxrd@z Thu Jan  1 00:00:00 1970
From: Jean-Christophe PLAGNIOL-VILLARD <plagnioj@jcrosoft.com>
Date: Mon, 19 May 2008 03:22:11 +0200
Subject: [U-Boot-Users] [PATCH 02/17] examples/eepro100_eeprom:
	Fix	memcpy to return destination pointer
In-Reply-To: <20080518230336.5DA772476E@gemini.denx.de>
References: <20080518222019.GC19480@game.jcrosoft.org>
	<20080518230336.5DA772476E@gemini.denx.de>
Message-ID: <20080519012211.GD19480@game.jcrosoft.org>
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To: u-boot@lists.denx.de

On 01:03 Mon 19 May     , Wolfgang Denk wrote:
> In message <20080518222019.GC19480@game.jcrosoft.org> you wrote:
> >
> > > >  static inline void *memcpy(void *dst, const void *src, unsigned int len)
> > > >  {
> > > >  	char *ret = dst;
> > > > +
> > > >  	while (len-- > 0) {
> > > >  		*ret++ = *((char *)src);
> > > >  		src++;
> > > >  	}
> > > > -	return (void *)ret;
> > > > +
> > > > +	return (void *)dst;
> > > 
> > > While technically correct, this is bogus. We have a variable ret, but
> > > we don't return it. And we have a variable dst, but we don't use it as
> > > destination pointer.
> > > 
> > > Please change the
> > > 	*ret++ = *((char *)src);
> > > into
> > > 	*dst++ = *((char *)src);
> > > and leave all the rest.
> > You can not do this because dst is a void
> 
> Why not? src is void, too.
gcc will claim about the cast.

we can do this
       void *ret = dst;
       char *d = dst;
       const char *s = src;

       while (len-- > 0)
               *d++ = *s++;

        return ret;

Best Regards,
J.