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Implement O(n) sort for inverted selection #1904

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Implement O(n) sort for inverted selection #1904

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48 changes: 40 additions & 8 deletions src/nnn.c
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Original file line number Diff line number Diff line change
Expand Up @@ -1724,12 +1724,40 @@ static char *findinsel(char *startpos, int len)
}
}

static int markcmp(const void *va, const void *vb)
{
const selmark *ma = (selmark*)va;
const selmark *mb = (selmark*)vb;
/*
* Sort selection markers by their starting position in O(n) time. Use one byte
* radix and eight passes (starting from the LSB).
*/
static void sort_marked(selmark *data, selmark *scratch, size_t nelem)
{
size_t cnt[256];
size_t idx[256];
size_t value;
size_t i, j;
void *tmp;
int b;
Comment on lines +1733 to +1738
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I'd prefer these names to be a bit more descriptive. What is b? cnt and idx make sense if you know what radix sort does, so that's more or less fine. i is an iteration variable that's fine, but j isn't, it's used for something completely unrelated.

Also small nitpick, we know the type of tmp, it should be selmark * just as data and scratch are.


for (b = 0; b < 8; b++) {
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The magic 8 being?

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@N-R-K N-R-K Jun 29, 2024 •
edited
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I had intended to comment on this. I think it's assuming sizeof(size_t) == 8, but that's not always true, nnn should work fine on 32bit systems too.

EDIT: just noticed the comment down below, looks like you figured it out :D

memset(cnt, 0, sizeof(cnt));
for (i = 0; i < nelem; i++) {
value = (size_t)data[i].startpos;
cnt[(value >> (b * 8)) & 0xFF]++;
}

return ma->startpos - mb->startpos;
idx[0] = 0;
for (i = 1; i < 256; i++)
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We mostly use ++i. Let's keep it consistent.

idx[i] = idx[i - 1] + cnt[i - 1];

for (i = 0; i < nelem; i++) {
value = (size_t)data[i].startpos;
j = idx[(value >> (b * 8)) & 0xFF]++;
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Only here do we learn what b actually is and where the magic 8 comes from. It's the index of the byte in value.

But this also assumes that size_t is actually 64-bit i.e. 8 bytes long. I think we support 32-bit and systems where size_t may not be 8 bytes.

Adjust the names and the limit for b to depend on the size of size_t.

memcpy(&scratch[j], &data[i], sizeof(selmark));
}

tmp = data;
data = scratch;
scratch = tmp;
}
}

/* scanselforpath() must be called before calling this */
Expand All @@ -1754,10 +1782,13 @@ static void invertselbuf(const int pathlen)
int i, nmarked = 0, prev = 0;
struct entry *dentp;
bool scan = FALSE;
selmark *marked = malloc(nselected * sizeof(selmark));
selmark *marked = calloc(nselected, sizeof(*marked));
selmark *scratch = calloc(nselected, sizeof(*scratch));
Comment on lines -1757 to +1786
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Why are we replacing the actual struct selmark with this dereferencing stuff? Is this even well-formed C? Initializing a pointer with a value that dereferences said pointer seems like an obvious no-no to me.

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Also, I just realized that it's changed to calloc(). Why? I don't think we need that.

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Is this even well-formed C? Initializing a pointer with a value that dereferences said pointer seems like an obvious no-no to me.

Yeah, that's fine. sizeof only needs to know the type of the operand in order to derive it's size. It's guaranteed not to evaluate anything (with the exception when the operand is a VLA, one more reason I hate VLAs).

For reference C11 6.5.3.4:

If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated

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The more you know. I'm still more for sizeof(selmark), using an identifier during its initialization just looks off to me. Up to you @jarkkojs.

calloc() still unnecessary and I'd still move scrath into the sorting function.


if (!marked) {
if (marked == NULL || scratch == NULL) {
printwarn(NULL);
free(marked);
free(scratch);
return;
}

Expand Down Expand Up @@ -1810,7 +1841,7 @@ static void invertselbuf(const int pathlen)
* With entries sorted we can merge adjacent ones allowing us to
* move them in a single go.
*/
qsort(marked, nmarked, sizeof(selmark), &markcmp);
sort_marked(marked, scratch, nselected);

/* Some files might be adjacent. Merge them into a single entry */
for (i = 1; i < nmarked; ++i) {
Expand Down Expand Up @@ -1850,6 +1881,7 @@ static void invertselbuf(const int pathlen)
}

free(marked);
free(scratch);
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Should scratch really be allocated here? It's only used in sort_marked() and I see no benefit in allocating it in this function. Might as well just allocate and free it there.


/* Buffer size adjustment */
selbufpos -= shrinklen;
Expand Down
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