ITAGAKI Takahiro wrote:
Heikki Linnakangas <[EMAIL PROTECTED]> wrote:
If I'm reading the code correctly, DSM makes no attempt to keep the chunks ordered by block number. If that's the case, vacuum needs to be modified because it currently relies on the fact that blocks are scanned and the dead tuple list is therefore populated in order.

Vacuum still scans heaps in block order and picks up corresponding DSM
chunks. Therefore the order of DSM chunks is not important. This method
is not efficient for huge tables with small deadspaces, but I think it
doesn't become a serious issue.

Ok, I can see now that the iterator returns pages in heap order, so no problem there.

Looking closer at FlushBuffer: before flushing a page to disk, the page is scanned to count the number of vacuumable tuples on it. That has the side effect of setting all the hint bits, which is something that we've been thinking of doing anyway (there's a TODO on that as well). It adds some CPU overhead to writing dirty buffers, which I personally don't believe is a problem at all, but it's worth noting. I don't believe it's a problem because if you're system is I/O bound, the CPU overhead doesn't really matter and if it saves any I/O later by not having to dirty the page later to write the hint bits, the benefit definitely outweights the cost. And if your system is CPU bound, there shouldn't be that many FlushBuffers happening for it to matter too much.

I think we should set the bit in the DSM whenever there's any dead space in the block, instead of having the threshold of 2 tuples or BLCKSZ/4 space. A pathological example is a relatively seldom updated table with a fillfactor set so that there's room for exactly one update on each page. The DSM will never have any bits set for the table, because there's no room for more than one dead tuple on any page.

But if we don't bother with the threshold, we don't need to scan the tuples in FlushBuffer to count the dead tuples. I think it'd still be worth it to scan them just to set the hint bits, though, but it becomes an orthogonal feature then.

  Heikki Linnakangas

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