Remove obsolete comment.

Commit 8b304b8b72b0a60f1968d39f01cf817c8df863ec removed replacement
selection, but left behind this comment text.  The optimization to
which the comment refers is not relevant without replacement
selection, because if we had so few tuples as to require only one
tape, we would have just completed the sort in memory.

Peter Geoghegan

Discussion: http://postgr.es/m/CAH2-WznqupLA8CMjp+vqzoe0yXu0DYYbQSNZxmgN76tLnAOZ_w@mail.gmail.com

diff --git a/src/backend/utils/sort/tuplesort.c b/src/backend/utils/sort/tuplesort.c
index 3c23ac75a09ca27e9aaec732a3285e7b5ecbd2f1..35eebad8e4c04c7bf17d586d519d1020c3dcaafd 100644 (file)
--- a/src/backend/utils/sort/tuplesort.c
+++ b/src/backend/utils/sort/tuplesort.c
@@ -2459,11 +2459,6 @@ mergeruns(Tuplesortstate *state)
         * Use all the remaining memory we have available for read buffers among
         * the input tapes.
         *
-        * We do this only after checking for the case that we produced only one
-        * initial run, because there is no need to use a large read buffer when
-        * we're reading from a single tape.  With one tape, the I/O pattern will
-        * be the same regardless of the buffer size.
-        *
         * We don't try to "rebalance" the memory among tapes, when we start a new
         * merge phase, even if some tapes are inactive in the new phase.  That
         * would be hard, because logtape.c doesn't know where one run ends and