如何避免批处理块中的重置超时

Question

I'm using TPL DataFlow with BatchBlock which is triggered automatically when the number of currently queued or postponed items is less than the BatchSize after a timeout as given below:

Timer triggerBatchTimer = new(_ => _batchBlock.TriggerBatch());

TransformBlock<string, string> timeoutTransformBlock = new((value) =>
{
    triggerBatchTimer.Change(_options.Value.TriggerBatchTime, Timeout.Infinite);
    return value;
});

var actionBlock = new ActionBlock<IEnumerable<string>>(action =>
{
    GenerateFile(action);
});

_buffer.LinkTo(timeoutTransformBlock);
timeoutTransformBlock.LinkTo(_batchBlock);
_batchBlock.LinkTo(actionBlock);

Ie with max batch size: 4 and timeout 10sec

Current behavior
BatchBlock (items): +---------1------------2------3---------------------------------|
Timeout (sec)     : +-10--9--10--9--8--7--10--9--10--9--8--7--6--5--4--3--2--1--0---|
ActionBlock       : +----------------------------------------------------------Call-|

Expected
BatchBlock (items): +--------1-----------2-----3---------|
Timeout (sec)     : +-10--9--8--7--6--5--4--3--2--1--0---|
ActionBlock       : +-------------------------------Call-|

But my problem is how to avoid timeout to be reset to 0 each time the block receive a new item?

Answer 1

Thanks to @Theodor Zoulias for help, this is solution:

Just initialize the timer with the same value for both dueTime and period and remove triggerBatchTimer.Change in TransformBlock . Thus batchblock fires every X seconds or when the batchSize has been reached and the timer is not reset for each new item in the batchblock.

Timer triggerBatchTimer = new(_ => _batchBlock.TriggerBatch(), null, options.Value.TriggerBatchTime, options.Value.TriggerBatchTime);

TransformBlock<string, string> timeoutTransformBlock = new((value) =>
{
    return value;
});

var actionBlock = new ActionBlock<IEnumerable<string>>(action =>
{
    GenerateFile(action);
});

_buffer.LinkTo(timeoutTransformBlock);
timeoutTransformBlock.LinkTo(_batchBlock);
_batchBlock.LinkTo(actionBlock);