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TCK: Enable replay-like and delay-error like Processor impls #348
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Original file line number | Diff line number | Diff line change |
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@@ -11,6 +11,8 @@ | |
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package org.reactivestreams.tck; | ||
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import java.util.*; | ||
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import org.reactivestreams.Processor; | ||
import org.reactivestreams.Publisher; | ||
import org.reactivestreams.Subscriber; | ||
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@@ -25,9 +27,6 @@ | |
import org.testng.annotations.BeforeMethod; | ||
import org.testng.annotations.Test; | ||
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import java.util.HashSet; | ||
import java.util.Set; | ||
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public abstract class IdentityProcessorVerification<T> extends WithHelperPublisher<T> | ||
implements SubscriberWhiteboxVerificationRules, PublisherVerificationRules { | ||
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@@ -189,6 +188,19 @@ public long maxSupportedSubscribers() { | |
return Long.MAX_VALUE; | ||
} | ||
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/** | ||
* Indicates that the tested implementation keeps a strict event ordering in respect to | ||
* {@code onNext} and {@code onError}. | ||
* Some {@code Processor} implementation may emit all {@code onNext}s received before emitting | ||
* any {@code onError}, similar to how {@code onComplete} is usually emitted after all | ||
* previous {@code onNext} items have been emitted. The default implementation returns false, | ||
* indicating that {@code onError} may cut ahead and get emitted even if there are | ||
* {@code onNext} events ready for consumption (via {@code request()}) by the {@code Subscriber}. | ||
*/ | ||
public boolean strictEventOrdering() { | ||
return false; | ||
} | ||
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////////////////////// TEST ENV CLEANUP ///////////////////////////////////// | ||
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@BeforeMethod | ||
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@@ -417,6 +429,14 @@ public TestSetup apply(Long aLong) throws Throwable { | |
final Exception ex = new RuntimeException("Test exception"); | ||
sendError(ex); | ||
sub1.expectError(ex); | ||
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// some Processors may only emit the terminal error if | ||
// all previously submitted onNext value has been | ||
// consumed by the Subscriber | ||
if (strictEventOrdering()) { | ||
sub2.request(1); | ||
expectNextElement(sub2, x); | ||
} | ||
sub2.expectError(ex); | ||
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env.verifyNoAsyncErrorsNoDelay(); | ||
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@@ -668,7 +688,10 @@ public TestSetup apply(Long subscribers) throws Throwable { | |
sub2.expectNone(); // since sub2 hasn't requested anything yet | ||
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sub2.request(1); | ||
expectNextElement(sub2, z); | ||
// some processors could emit items from the beginning, some may | ||
// cache for a limited time or count, therefore | ||
// any of the first 3 items may appear after requesing one | ||
expectAnyNextElement(sub2, new HashSet<T>(Arrays.asList(x, y, z))); | ||
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if (totalRequests == 3) { | ||
expectRequest(); | ||
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@@ -677,7 +700,20 @@ public TestSetup apply(Long subscribers) throws Throwable { | |
// to avoid error messages during test harness shutdown | ||
sendCompletion(); | ||
sub1.expectCompletion(env.defaultTimeoutMillis()); | ||
sub2.expectCompletion(env.defaultTimeoutMillis()); | ||
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// sub2 may not complete because it still has pending y or z | ||
if (!sub2.tryExpectCompletion(env.defaultTimeoutMillis())) { | ||
sub2.request(1); | ||
expectAnyNextElement(sub2, new HashSet<T>(Arrays.asList(y, z))); | ||
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// z may still be pending | ||
if (!sub2.tryExpectCompletion(env.defaultTimeoutMillis())) { | ||
sub2.request(1); | ||
expectNextElement(sub2, z); | ||
// after z, it should get onComplete reasonably quickly | ||
sub2.expectCompletion(env.defaultTimeoutMillis()); | ||
} | ||
} | ||
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env.verifyNoAsyncErrorsNoDelay(); | ||
}}; | ||
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@@ -738,6 +774,20 @@ public void expectNextElement(ManualSubscriber<T> sub, T expected) throws Interr | |
} | ||
} | ||
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public void expectAnyNextElement(ManualSubscriber<T> sub, Set<T> expected) throws InterruptedException { | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Updated. |
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final T elem = sub.nextElement(String.format("timeout while awaiting %s", expected)); | ||
if (!expected.contains(elem)) { | ||
StringBuilder sb = new StringBuilder(); | ||
for (T t : expected) { | ||
if (sb.length() > 0) { | ||
sb.append(", "); | ||
} | ||
sb.append(String.format("`onNext(%s)`", t)); | ||
} | ||
env.flop(String.format("Received `onNext(%s)` on downstream but expected any of %s", elem, sb)); | ||
} | ||
} | ||
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public T sendNextTFromUpstream() throws InterruptedException { | ||
final T x = nextT(); | ||
sendNext(x); | ||
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@@ -506,6 +506,10 @@ public void expectCompletion(long timeoutMillis, String errorMsg) throws Interru | |
received.expectCompletion(timeoutMillis, errorMsg); | ||
} | ||
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public boolean tryExpectCompletion(long timeoutMillis) throws InterruptedException { | ||
return received.tryExpectCompletion(timeoutMillis); | ||
} | ||
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public <E extends Throwable> void expectErrorWithMessage(Class<E> expected, String requiredMessagePart) throws Exception { | ||
expectErrorWithMessage(expected, requiredMessagePart, env.defaultTimeoutMillis()); | ||
} | ||
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@@ -973,6 +977,22 @@ public void expectCompletion(long timeoutMillis, String errorMsg) throws Interru | |
} // else, ok | ||
} | ||
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public boolean tryExpectCompletion(long timeoutMillis) throws InterruptedException { | ||
long end = System.currentTimeMillis() + timeoutMillis; | ||
do { | ||
Optional<T> value = abq.peek(); | ||
if (value != null) { | ||
if (!value.isDefined()) { | ||
abq.poll(); | ||
return true; | ||
} | ||
return false; | ||
} | ||
Thread.sleep(1); | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Why this ISO poll(remaining)? There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. I don't understand this question. This routine uses peek because in case the next received element is not a terminal indicator, that element should remain available for any of the other There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. @akarnokd if you use abq.poll(remaining) instead of abq.peek() + abp.poll() you do not need to do the sleep(1)-dance. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more.
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 1: https://docs.oracle.com/javase/7/docs/api/java/util/concurrent/ArrayBlockingQueue.html#poll(long,%20java.util.concurrent.TimeUnit) |
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} while (System.currentTimeMillis() < end); | ||
return false; | ||
} | ||
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@SuppressWarnings("unchecked") | ||
public <E extends Throwable> E expectError(Class<E> clazz, long timeoutMillis, String errorMsg) throws Exception { | ||
Thread.sleep(timeoutMillis); | ||
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I'm n ot super-convinced by the name. Isn't this more about
eagerTermination
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or
shortCircuitTermination
?