Please correct me, but as best I can tell, the Enum.map will, sequentially, start the first receive, and that receive will operate against every incoming messag

Please correct me, but as best I can tell, the Enum.map will, sequentially, start the first receive, and that receive will operate against every incoming message until ^pid matches, adding its result as the first member of the collection while the unmatched messages accumulate in the mailbox; then it will do the second receive, which will scan the mailbox for its ^pid, and either find it or wait until the matching message arrives, adding its result as the second member of the collection; then it will do the third receive, etc. Assuming a random distribution of message arrivals, the last message to arrive will on average be the n/2 message, so for (on average) n/2 messages, the n/2 length mailbox (full of the (n/2) through n messages) will need to be scanned n/2 times, e.g., O(n*n). There might be some way to reduce the mailbox scanning time to O(1), but my shallow dive in the source code didn't show it.