Re: [PATCH 1/8] lib/sort: Heapsort implementation of sort()

From: linux-os
Date: Tue Feb 01 2005 - 13:18:03 EST


On Tue, 1 Feb 2005, Andreas Gruenbacher wrote:

On Mon, 2005-01-31 at 18:30, Paulo Marques wrote:
Andreas Gruenbacher wrote:
[...]

static inline void swap(void *a, void *b, int size)
{
if (size % sizeof(long)) {
char t;
do {
t = *(char *)a;
*(char *)a++ = *(char *)b;
*(char *)b++ = t;
} while (--size > 0);
} else {
long t;
do {
t = *(long *)a;
*(long *)a = *(long *)b;
*(long *)b = t;
size -= sizeof(long);
} while (size > sizeof(long));

You forgot to increment a and b, and this should be "while (size);", no?

Correct, yes.

Or better yet,

static inline void swap(void *a, void *b, int size)
{
long tl;
char t;

while (size >= sizeof(long)) {
tl = *(long *)a;
*(long *)a = *(long *)b;
*(long *)b = tl;
a += sizeof(long);
b += sizeof(long);
size -= sizeof(long);
}
while (size) {
t = *(char *)a;
*(char *)a++ = *(char *)b;
*(char *)b++ = t;
size--;
}
}

This works better if the size is not a multiple of sizeof(long), but is
bigger than a long.

In case size is not a multiple of sizeof(long) here, you'll have
unaligned memory accesses most of the time anyway, so it probably
doesn't really matter.

Thanks,
--
Andreas Gruenbacher <agruen@xxxxxxx>
SUSE Labs, SUSE LINUX GMBH

This uses an GNU-ISM where you are doing pointer arithmetic
on a pointer-to-void. NotGood(tm) this is an example of
where there is absolutely no rationale whatsoever for
writing such code.


Cheers,
Dick Johnson
Penguin : Linux version 2.6.10 on an i686 machine (5537.79 BogoMips).
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