memcpy

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

Am Donnerstag 27 August 2009 00:18:28 schrieb Stoyan Gaydarov:

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

It is IMHO agreed concensus that kernel-implementations of well-known (and especially C-library) userspace functions have to behave the same/similar.
Or they get another name if only to avoid confusion.

Use the source, Luke.

Bernd

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