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go - 对这些 golang 并发问题感到困惑

问题描述

我刚刚完成了 coursera ( https://www.coursera.org/learn/golang-concurrency/ ) 上的“Go 并发”课程，最后的作业我真的很挣扎。这是我的提交：

// 1.There should be 5 philosophers sharing chopsticks, with one chopstick between each adjacent pair of philosophers.
// 2.Each philosopher should eat only 3 times (not in an infinite loop as we did in lecture)
// 3.The philosophers pick up the chopsticks in any order, not lowest-numbered first (which we did in lecture).
// 4.In order to eat, a philosopher must get permission from a host which executes in its own goroutine.
// 5.The host allows no more than 2 philosophers to eat concurrently.
// 6.Each philosopher is numbered, 1 through 5.
// 7.When a philosopher starts eating (after it has obtained necessary locks) it prints “starting to eat <number>” on a line by itself, where <number> is the number of the philosopher.
// 8.When a philosopher finishes eating (before it has released its locks) it prints “finishing eating <number>” on a line by itself, where <number> is the number of the philosopher.
package main

import (
    "fmt"
    "sync"
)

var eating = make(chan int, 2)

var mu sync.RWMutex
var everyoneAte int
var timesEaten = make(map[int]int, 5)

type chopstick struct {
    sync.Mutex
}

type philosopher struct {
    leftCs  *chopstick
    rightCs *chopstick
}

func alreadyAte(index int) bool {
    mu.Lock()
    defer mu.Unlock()
    if timesEaten[index] == 3 {
        return true
    }
    return false
}
func (p philosopher) eat(index int) {

    eating <- 1

    p.leftCs.Lock()
    p.rightCs.Lock()

    fmt.Printf("Starting to eat %v\n", index)
    fmt.Printf("Finishing eating %v\n", index)
    mu.Lock()
    timesEaten[index]++
    if timesEaten[index] == 3 {
        everyoneAte++
    }
    mu.Unlock()
    p.rightCs.Unlock()
    p.leftCs.Unlock()
    <-eating
}

func main() {
    count := 5
    chopsticks := make([]*chopstick, count)
    for i := 0; i < count; i++ {
        chopsticks[i] = &chopstick{}
    }

    philosophers := make([]*philosopher, count)
    for i := 0; i < count; i++ {
        philosophers[i] = &philosopher{
            leftCs:  chopsticks[i],
            rightCs: chopsticks[(i+1)%count],
        }
    }
    for {
        mu.RLock()
        if everyoneAte == count {
            return
        }
        for i := 0; i < count; i++ {
            if timesEaten[i] == 3 {
                continue
            }
            go philosophers[i].eat(i + 1)
        }
        mu.RUnlock()
    }

}

我不明白如何实现#4，所以我只使用了缓冲通道 ‍♂️</p>

不明白为什么有些哲人回来后吃了3次以上

如果每个人都有这些问题的答案，我将不胜感激。我已经提交了作业。

标签： goconcurrency

解决方案

这是哲学家可能吃三遍以上的顺序

假设哲学家_1 已经吃过 2 次

  1. 主要：获取RLock
  2. 主要：读timesEaten[1] == 2
  3. main:philosopher_1在一个单独的 goroutine_1 上做吃
  4. 主要：发布RLock
  5. 主要：获取RLock
  6. 主要：读timesEaten[1] == 2
  7. main:philosopher_1在一个单独的 goroutine_2 上再次进食
  8. 主要：发布RLock
  9. goroutine_1：获取Lock
  10. goroutine_1：设置timesEaten[1] = 3
  11. goroutine_1：发布Lock
  12. goroutine_2：获取Lock
  13. goroutine_2: 设置timesEaten[1] = 4<--philosopher_1吃了超过 3 次
  14. goroutine_2：发布Lock

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