Fix 'request(0)' issue in Scan #1957
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| @@ -147,19 +147,17 @@ public void request(long n) { | |||
| if (once.compareAndSet(false, true)) { | |||
| if (initialValue == NO_INITIAL_VALUE || n == Long.MAX_VALUE) { | |||
| producer.request(n); | |||
| } else if (n == 1) { | |||
| excessive.set(true); | |||
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What does excessive mean? I'm re-reading this code and the intent of that variable doesn't jump out at me.
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It means we request more than what we should.
If a user requests 1 from scan in onStart, scan need to request 1 to the upstream. However, when scan receives an onNext, it will emit 2 onNext to the downstream: the initial value and accumulator.call(this.value, initialvalue). So the downstream only requests 1, but scan emits 2 values. I use excessive to indicate we are in an excessive state.
So the next time when the downstream requests, I will try to request n-1.
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If a user requests X (X > 1) from scan in onStart, scan can simply request (X-1). In such case, excessive will be false.
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Got it, thanks for the explanation.
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The logic seems correct to me on a re-read of this code. |
Fix 'request(0)' issue in Scan
#1904 didn't all cases.