| // Copyright 2017 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 semaphore_test | |
| import ( | |
| "context" | |
| "fmt" | |
| "log" | |
| "runtime" | |
| "golang.org/x/sync/semaphore" | |
| ) | |
| // Example_workerPool demonstrates how to use a semaphore to limit the number of | |
| // goroutines working on parallel tasks. | |
| // | |
| // This use of a semaphore mimics a typical “worker pool” pattern, but without | |
| // the need to explicitly shut down idle workers when the work is done. | |
| func Example_workerPool() { | |
| ctx := context.TODO() | |
| var ( | |
| maxWorkers = runtime.GOMAXPROCS(0) | |
| sem = semaphore.NewWeighted(int64(maxWorkers)) | |
| out = make([]int, 32) | |
| ) | |
| // Compute the output using up to maxWorkers goroutines at a time. | |
| for i := range out { | |
| // When maxWorkers goroutines are in flight, Acquire blocks until one of the | |
| // workers finishes. | |
| if err := sem.Acquire(ctx, 1); err != nil { | |
| log.Printf("Failed to acquire semaphore: %v", err) | |
| break | |
| } | |
| go func(i int) { | |
| defer sem.Release(1) | |
| out[i] = collatzSteps(i + 1) | |
| }(i) | |
| } | |
| // Acquire all of the tokens to wait for any remaining workers to finish. | |
| // | |
| // If you are already waiting for the workers by some other means (such as an | |
| // errgroup.Group), you can omit this final Acquire call. | |
| if err := sem.Acquire(ctx, int64(maxWorkers)); err != nil { | |
| log.Printf("Failed to acquire semaphore: %v", err) | |
| } | |
| fmt.Println(out) | |
| // Output: | |
| // [0 1 7 2 5 8 16 3 19 6 14 9 9 17 17 4 12 20 20 7 7 15 15 10 23 10 111 18 18 18 106 5] | |
| } | |
| // collatzSteps computes the number of steps to reach 1 under the Collatz | |
| // conjecture. (See https://en.wikipedia.org/wiki/Collatz_conjecture.) | |
| func collatzSteps(n int) (steps int) { | |
| if n <= 0 { | |
| panic("nonpositive input") | |
| } | |
| for ; n > 1; steps++ { | |
| if steps < 0 { | |
| panic("too many steps") | |
| } | |
| if n%2 == 0 { | |
| n /= 2 | |
| continue | |
| } | |
| const maxInt = int(^uint(0) >> 1) | |
| if n > (maxInt-1)/3 { | |
| panic("overflow") | |
| } | |
| n = 3*n + 1 | |
| } | |
| return steps | |
| } |