| // Copyright 2019 The Go Authors. All rights reserved. | |
| // Use of this source code is governed by a BSD-style | |
| // license that can be found in the LICENSE file. | |
| package runtime_test | |
| import ( | |
| "fmt" | |
| . "runtime" | |
| "sync" | |
| "sync/atomic" | |
| "testing" | |
| ) | |
| // TestSemaHandoff checks that when semrelease+handoff is | |
| // requested, the G that releases the semaphore yields its | |
| // P directly to the first waiter in line. | |
| // See issue 33747 for discussion. | |
| func TestSemaHandoff(t *testing.T) { | |
| const iter = 10000 | |
| ok := 0 | |
| for i := 0; i < iter; i++ { | |
| if testSemaHandoff() { | |
| ok++ | |
| } | |
| } | |
| // As long as two thirds of handoffs are direct, we | |
| // consider the test successful. The scheduler is | |
| // nondeterministic, so this test checks that we get the | |
| // desired outcome in a significant majority of cases. | |
| // The actual ratio of direct handoffs is much higher | |
| // (>90%) but we use a lower threshold to minimize the | |
| // chances that unrelated changes in the runtime will | |
| // cause the test to fail or become flaky. | |
| if ok < iter*2/3 { | |
| t.Fatal("direct handoff < 2/3:", ok, iter) | |
| } | |
| } | |
| func TestSemaHandoff1(t *testing.T) { | |
| if GOMAXPROCS(-1) <= 1 { | |
| t.Skip("GOMAXPROCS <= 1") | |
| } | |
| defer GOMAXPROCS(GOMAXPROCS(-1)) | |
| GOMAXPROCS(1) | |
| TestSemaHandoff(t) | |
| } | |
| func TestSemaHandoff2(t *testing.T) { | |
| if GOMAXPROCS(-1) <= 2 { | |
| t.Skip("GOMAXPROCS <= 2") | |
| } | |
| defer GOMAXPROCS(GOMAXPROCS(-1)) | |
| GOMAXPROCS(2) | |
| TestSemaHandoff(t) | |
| } | |
| func testSemaHandoff() bool { | |
| var sema, res uint32 | |
| done := make(chan struct{}) | |
| // We're testing that the current goroutine is able to yield its time slice | |
| // to another goroutine. Stop the current goroutine from migrating to | |
| // another CPU where it can win the race (and appear to have not yielded) by | |
| // keeping the CPUs slightly busy. | |
| var wg sync.WaitGroup | |
| for i := 0; i < GOMAXPROCS(-1); i++ { | |
| wg.Add(1) | |
| go func() { | |
| defer wg.Done() | |
| for { | |
| select { | |
| case <-done: | |
| return | |
| default: | |
| } | |
| Gosched() | |
| } | |
| }() | |
| } | |
| wg.Add(1) | |
| go func() { | |
| defer wg.Done() | |
| Semacquire(&sema) | |
| atomic.CompareAndSwapUint32(&res, 0, 1) | |
| Semrelease1(&sema, true, 0) | |
| close(done) | |
| }() | |
| for SemNwait(&sema) == 0 { | |
| Gosched() // wait for goroutine to block in Semacquire | |
| } | |
| // The crux of the test: we release the semaphore with handoff | |
| // and immediately perform a CAS both here and in the waiter; we | |
| // want the CAS in the waiter to execute first. | |
| Semrelease1(&sema, true, 0) | |
| atomic.CompareAndSwapUint32(&res, 0, 2) | |
| wg.Wait() // wait for goroutines to finish to avoid data races | |
| return res == 1 // did the waiter run first? | |
| } | |
| func BenchmarkSemTable(b *testing.B) { | |
| for _, n := range []int{1000, 2000, 4000, 8000} { | |
| b.Run(fmt.Sprintf("OneAddrCollision/n=%d", n), func(b *testing.B) { | |
| tab := Escape(new(SemTable)) | |
| u := make([]uint32, SemTableSize+1) | |
| b.ResetTimer() | |
| for j := 0; j < b.N; j++ { | |
| // Simulate two locks colliding on the same semaRoot. | |
| // | |
| // Specifically enqueue all the waiters for the first lock, | |
| // then all the waiters for the second lock. | |
| // | |
| // Then, dequeue all the waiters from the first lock, then | |
| // the second. | |
| // | |
| // Each enqueue/dequeue operation should be O(1), because | |
| // there are exactly 2 locks. This could be O(n) if all | |
| // the waiters for both locks are on the same list, as it | |
| // once was. | |
| for i := 0; i < n; i++ { | |
| if i < n/2 { | |
| tab.Enqueue(&u[0]) | |
| } else { | |
| tab.Enqueue(&u[SemTableSize]) | |
| } | |
| } | |
| for i := 0; i < n; i++ { | |
| var ok bool | |
| if i < n/2 { | |
| ok = tab.Dequeue(&u[0]) | |
| } else { | |
| ok = tab.Dequeue(&u[SemTableSize]) | |
| } | |
| if !ok { | |
| b.Fatal("failed to dequeue") | |
| } | |
| } | |
| } | |
| }) | |
| b.Run(fmt.Sprintf("ManyAddrCollision/n=%d", n), func(b *testing.B) { | |
| tab := Escape(new(SemTable)) | |
| u := make([]uint32, n*SemTableSize) | |
| b.ResetTimer() | |
| for j := 0; j < b.N; j++ { | |
| // Simulate n locks colliding on the same semaRoot. | |
| // | |
| // Each enqueue/dequeue operation should be O(log n), because | |
| // each semaRoot is a tree. This could be O(n) if it was | |
| // some simpler data structure. | |
| for i := 0; i < n; i++ { | |
| tab.Enqueue(&u[i*SemTableSize]) | |
| } | |
| for i := 0; i < n; i++ { | |
| if !tab.Dequeue(&u[i*SemTableSize]) { | |
| b.Fatal("failed to dequeue") | |
| } | |
| } | |
| } | |
| }) | |
| } | |
| } |