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Date: Fri, 08 Jun 2012 10:30:02 -0400
From: pete <pfiland@mindspring.com>
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Subject: Re: copy size
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David T. Ashley wrote:
> 
> On Thu, 07 Jun 2012 22:40:52 -0700, Keith Thompson <kst-u@mib.org>
> wrote:
> 
> >
> >So yes, it's ok to call memcpy() with a size of 0.
> >
> >Note that if either s1 or s2 is a null pointer, the behavior is (I
> >believe) undefined.
> 
> I quick search of the Internet has revealed that you are right.
> 
> But it seems odd for three reasons:
> 
> a)In almost all the functions I write that have parameters of type
> (void *) and size_t (usually occurring together, for the obvious
> reasons), I handle the size=0 case.  When I handle it, the most
> natural construction of the function is that the pointer(s) are not
> dereferenced if size=0.

memcpy is of the <string.h> functions.
and there are some of them, 
where it may make sense to derefence a pointer parameter
prior to evaluating the n parameter.

char *strncat(char *s1, const char *s2, size_t n)
{
    char *const p1 = s1;

    s1 += str_len(s1);
    while (n-- != 0 && *s2 != '\0') {
        *s1++ = *s2++;
    }
    *s1 = '\0';
    return p1;
}

> 
> b)Allowing memcpy(dptr, sptr, 0) means that you decided to push logic
> to handle the size=0 case down into the function, by definition. 

memcpy may or may not, (depending on how it is implemented),
require any special code to handle the n equals 0 case.

void *memcpy(void *s1, const void *s2, size_t n)
{
          unsigned char *p1 = s1;
    const unsigned char *p2 = s2;

    while (n-- != 0) {
        *p1++ = *p2++;
    }
    return s1;
}

-- 
pete