From mboxrd@z Thu Jan 1 00:00:00 1970 From: Peter =?iso-8859-1?q?H=FCwe?= Date: Wed, 26 Aug 2009 22:42:00 +0000 Subject: Re: memcpy Message-Id: <200908270042.00597.PeterHuewe@gmx.de> List-Id: References: <4A95B4B4.3020408@uiuc.edu> In-Reply-To: <4A95B4B4.3020408@uiuc.edu> MIME-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit To: kernel-janitors@vger.kernel.org Am Donnerstag 27 August 2009 00:18:28 schrieb Stoyan Gaydarov: > I wanted to know what memcpy returned as a result, and if it > needs/should be checked. There are several places in the kernel where i > noticed it being used but i also saw a warning about the result not > being used, so i wanted to know a little more about it. > > -Stoyan > -- > To unsubscribe from this list: send the line "unsubscribe kernel-janitors" > in the body of a message to majordomo@vger.kernel.org > More majordomo info at http://vger.kernel.org/majordomo-info.html Hi, *memcpy is defined in .../lib/string.c (+ some macros etc - use cscope to find more references) as /** * memcpy - Copy one area of memory to another * @dest: Where to copy to * @src: Where to copy from * @count: The size of the area. * * You should not use this function to access IO space, use memcpy_toio() * or memcpy_fromio() instead. */ void *memcpy(void *dest, const void *src, size_t count) { char *tmp = dest; const char *s = src; while (count--) *tmp++ = *s++; return dest; } You can see it just returns the destination address that you have passed to it -> so I guess unless it is possible that you have already passed it a null pointer (which might horribly fail) it is not really necessary to check the result. Peter