Package semaphore

▾ Example (WorkerPool)

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.

Code:

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
}