Re: Implementing my own memcpy

Rajan wrote:
>
> Hi Netocrat ,
> With your code fragment, I find that you are using unsigned char* , but
> I may have to copy a structure.
> We certainly cannot do *dest++ = *src++ ; if they are void pointers.
> So how do we deal with this?

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The point about memcpy is that it copies memory areas, measured in
bytes, with no regard for the types involved. I gather you wish
the same thing, but with automatic creation of the destination. In
this case you need a different signature, such as:

void *dupmem(void *src, size_t sz);

The void * type can point at arbitrary things, and a size_t can
specify a size on any machine. But to use void* you have to
convert to other types, thus:

void *dupmem(void *src, size_t sz)
{
unsigned char *sp = src;
unsigned char *dst;

if (dst = malloc(sz)) /* memory is available */
while (sz--) *dst++ = *sp++; /* copy away */
return dst; /* will be NULL for failure */
} /* dupmem, untested */

Note how src is typed into sp, without any casts. Similarly the
reverse typing for the return value of dupmem. The usage will be,
for p some type of pointer:

if (p = dupmem(whatever, howbig)) {
/* success, carry on */
}
else {
/* abject failure, panic */
}

--
"I conclude that there are two ways of constructing a software
design: One way is to make it so simple that there are obviously
no deficiencies and the other way is to make it so complicated
that there are no obvious deficiencies." -- C. A. R. Hoare


.

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