foo

chunkify(InList, BaseChunkSize) ->
    %%io:format("length of list is ~p and chunk size is ~p ~n",[length(InList),BaseChunkSize]),
    case byte_size(term_to_binary(InList)) of
    Size when Size > BaseChunkSize ->
        NumberOfChunksLikely = ((Size div BaseChunkSize) + 1),
        ChunkThreshold = Size div NumberOfChunksLikely,
        chunkify(InList, ChunkThreshold, [], 0, []);
    _Else ->
        [InList]
    end.

chunkify([], _ChunkThreshold, [], 0, OutputChunks) ->
    lists:reverse(OutputChunks);
chunkify([], _ChunkThreshold, OutList, _OutListSize, OutputChunks) ->
    lists:reverse([lists:reverse(OutList) | OutputChunks]);
chunkify([InElement | RestInList], ChunkThreshold, OutList, OutListSize, OutputChunks) ->
    case byte_size(term_to_binary(InElement)) of
    Size when (Size + OutListSize) > ChunkThreshold andalso OutList /= []
             andalso length(OutList) > 9 ->
        {Left, Right} = lists:split(length(OutList)/2, OutList),
        chunkify(RestInList, ChunkThreshold, [], 0, [lists:reverse([InElement | Right]) | lists:reverse(Left, OutputChunks)]);            
    Size ->
        chunkify(RestInList, ChunkThreshold, [InElement | OutList], OutListSize + Size, OutputChunks)
    end.

I need a second cup of coffee to unravel all these reversals

I know. It's confusing. But it's also maybe next to useless with an append-only strategy. We have to rewrite the node(s) anyway on each write whether chunkify makes one or multiple chunks. On the other hand, it _might_ have a positive effect on disk usage since the nodes might tend to remain smaller for longer. On Jul 1, 2011 4:02 AM, "bdionne" < reply@reply.github.com> wrote: