Go 中的哲学家用餐问题未通过单元测试

Question

我正在学习围棋课程，作业如下：

Implement the dining philosopher's problem with the following constraints/modifications.

There should be 5 philosophers sharing chopsticks, with one chopstick between each adjacent pair of philosophers.

Each philosopher should eat only 3 times (not in an infinite loop as we did in lecture).

The philosophers pick up the chopsticks in any order, not lowest-numbered first (which we did in lecture).

In order to eat, a philosopher must get permission from a host which executes in its own goroutine.

The host allows no more than 2 philosophers to eat concurrently.

Each philosopher is numbered, 1 through 5.

When a philosopher starts eating (after it has obtained necessary locks) it prints "starting to eat " on a line by itself, where is the number of the philosopher.

When a philosopher finishes eating (before it has released its locks) it prints "finishing eating " on a line by itself, where is the number of the philosopher.

我的实现：

package main

import (
    "fmt"
    "io"
    "math/rand"
    "os"
    "sync"
    "time"
)

const (
    NumPhilosophers    = 5
    NumEatMaxTimes     = 3
    NumMaxAllowedToEat = 2
)

type chopstick struct{ sync.Mutex }

type philosopher struct {
    num int
    cs  []*chopstick
}

func setTable() []*philosopher {
    cs := make([]*chopstick, NumPhilosophers)
    for i := 0; i < NumPhilosophers; i++ {
        cs[i] = new(chopstick)
    }
    ph := make([]*philosopher, NumPhilosophers)
    for i := 0; i < NumPhilosophers; i++ {
        ph[i] = &philosopher{i + 1, []*chopstick{cs[i], cs[(i+1)%NumPhilosophers]}}
    }

    return ph
}

func (ph philosopher) eat(sem chan int, wg *sync.WaitGroup, w io.Writer) {
    for i := 0; i < NumEatMaxTimes; i++ {
        /* Ask host for permission to eat */
        sem <- 1
        /*
            Pick any of the left or right chopsticks.
            Notice how the methods on the Mutex can be called directly on a chopstick due to embedding.
        */
        firstCS := rand.Intn(2)
        secondCS := (firstCS + 1) % 2
        ph.cs[firstCS].Lock()
        ph.cs[secondCS].Lock()

        fmt.Fprintf(w, "Starting to eat %d\n", ph.num)
        x := rand.Intn(NumEatMaxTimes)
        time.Sleep(time.Duration(x) * time.Second)
        fmt.Fprintf(w, "Finishing eating %d\n", ph.num)

        ph.cs[secondCS].Unlock()
        ph.cs[firstCS].Unlock()
        <-sem
    }
    wg.Done()
}

func main() {
    run(os.Stdout)
}

func run(w io.Writer) {
    var sem = make(chan int, NumMaxAllowedToEat)
    rand.Seed(time.Now().UnixNano())
    var wg sync.WaitGroup

    allPh := setTable()
    wg.Add(len(allPh))
    for _, ph := range allPh {
        go ph.eat(sem, &wg, w)
    }
    wg.Wait()
}

单元测试：

func TestRun(t *testing.T) {
    var out bytes.Buffer
    run(&out)
    lines := strings.Split(strings.ReplaceAll(out.String(), "\r\n", "\n"), "\n")
    eating := make(map[int]bool)
    timesEaten := make(map[int]int)
    for _, line := range lines {
        if line == "" {
            continue
        }
        fmt.Println(line)
        tokens := strings.Fields(line)

        i, err := strconv.Atoi(tokens[len(tokens)-1])
        if err != nil {
            t.Errorf("Bad line: %s", line)
        }

        s := strings.ToLower(tokens[0])

        if s == "starting" {
            if len(eating) > (NumMaxAllowedToEat - 1) {
                t.Errorf("%v are eating at the same time", eating)
            }
            _, ok := eating[i]
            if ok {
                t.Errorf("%d started before finishing", i)
            }
            eating[i] = true
        } else if s == "finishing" {
            _, ok := eating[i]
            if !ok {
                t.Errorf("%d finished without starting", i)
            }

            delete(eating, i)

            timesEaten[i] = timesEaten[i] + 1
        }
    }

    for k, v := range timesEaten {
        if v > NumEatMaxTimes {
            t.Errorf("%d ate %d times", k, v)
        }
    }

    if len(timesEaten) != NumPhilosophers {
        t.Error("One or more didn't get to eat")
    }
}

问题是，测试随机失败。下面是一个执行（添加了行号）：

1. Starting to eat 5
2. Starting to eat 2
3. Finishing eating 2
4. Finishing eating 5
5. Starting to eat 3
6. Starting to eat 1
7. Finishing eating 1
8. Finishing eating 3
9. Starting to eat 2
10. Starting to eat 4
11. Finishing eating 4
12. Starting to eat 5
13. Finishing eating 2
14. Finishing eating 5
15. Starting to eat 3
16. Finishing eating 3
17. Starting to eat 1
18. Finishing eating 4
19. Finishing eating 1
20. Starting to eat 5
21. Finishing eating 5
22. Starting to eat 3
23. Finishing eating 3
24. Starting to eat 4
25. Starting to eat 2
26. Finishing eating 2
27. Starting to eat 1
28. Finishing eating 4
29. Finishing eating 1

--- FAIL: TestRun (12.01s)
    main_test.go:43: 4 finished without starting

哲学家 4 从第 10 和 24 行开始，在第 11、18 和 28 行结束。第 28 行不匹配，因此测试正确报错。但是，我很难找到错误。你能帮忙吗？

Answer 1

回答我自己的问题，结果 byes.Buffer is not thread-safe. I ended up using go-fakeio 用于测试的库如下所示。

s, err := fakeio.Stderr().Stdout().Do(run)
if err != nil {
    t.Errorf("%v", err)
}

测试的其余部分保持不变。 main.run 函数不再需要 io.Writer，因为 fakeio 库替换了 stderr 和 stdout。

Go 中的哲学家用餐问题未通过单元测试

Dining philosophers problem in Go fails unit test

multithreading

synchronization

go

dining-philosopher

goroutine