/**
* Converts the bin/index slot, whether a singleton LN or a DIN root.
*
* <p>Enter/leave with bin field latched, although bin field may change.
*
* <p>When a singleton LN is converted, leaves with bin/index fields unchanged.
*
* <p>When a dup tree is converted, leaves with bin/index fields set to last inserted slot. This
* is the slot of the highest key in the dup tree.
*/
private void convertBinSlot() {
if (DEBUG) {
System.out.println(
"DupConvert BIN LSN "
+ DbLsn.getNoFormatString(bin.getLsn(index))
+ " index "
+ index
+ " nEntries "
+ bin.getNEntries());
}
/* Delete slot if LN is deleted. */
final boolean isDeleted;
if (isLNDeleted(bin, index)) {
deleteSlot();
return;
}
final Node node = bin.fetchLNOrDIN(index, CacheMode.DEFAULT);
if (!node.containsDuplicates()) {
if (DEBUG) {
System.out.println("DupConvert BIN LN " + Key.dumpString(bin.getKey(index), 0));
}
/* Fetching a non-deleted LN updates the slot key; we're done. */
assert node instanceof LN;
nConverted += 1;
return;
}
/*
* Delete the slot containing the DIN before re-inserting the dup tree,
* so that the DIN slot key doesn't interfere with insertions.
*
* The DIN is evicted and memory usage is decremented. This is not
* exactly correct because we keep a local reference to the DIN until
* the dup tree is converted, but we tolerate this temporary
* inaccuracy.
*/
final byte[] binKey = bin.getKey(index);
final DIN din = (DIN) node;
deleteSlot();
convertDin(din, binKey);
}
