同时被这些golang问题所迷惑

我刚刚在coursera（）上完成了“go中的并发性”课程，最后的作业让我很为难。这是我的意见：

// 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()
    }

}

//1.应该有5位哲学家共用筷子，每对相邻的哲学家共用一根筷子。
//2.每个哲学家只应该吃3次（而不是像我们在讲座中那样无限循环）
//3.哲学家们拿起筷子的顺序是任意的，而不是排在最前面（我们在课堂上就是这么做的）。
//为了吃饭，哲学家必须得到在自己的goroutine中执行的主机的许可。
//5.主人允许不超过2位哲学家同时用餐。
//每个哲学家都有编号，从1到5。
//当哲学家开始吃饭时（在获得必要的锁后），它会自己在一行上打印“开始吃饭”，其中是哲学家的编号。
//当一个哲学家吃完饭（在他解开锁之前），他会自己在一行上打印“吃完”，其中是哲学家的编号。
包干管
进口(
“fmt”
“同步”
)
变量吃=制造（chan int，2）
var mu sync.RWMutex
变量everyoneAte int
var timeseat=make（映射[int]int，5）
型筷子结构{
同步互斥
}
类型结构{
左撇子*筷子
右筷子
}
func alreadyAte（索引int）布尔{
木锁()
延迟mu.Unlock（）
如果时间消耗[索引]==3{
返回真值
}
返回错误
}
func（p）eat（索引int）{
吃这是哲学家可能吃三次以上的顺序

假设哲学家_1吃了2次

主：获取
RLock
main:reads
timeseat[1]==2
主要功能：让哲学家在一个单独的餐桌上吃饭
主：释放
RLock
主：获取
RLock
main:reads
timeseat[1]==2
主要功能：让哲学家在一个单独的餐桌上再吃一顿
主：释放
RLock
goroutine_1：获取
锁
goroutine_1:set
timeseat[1]=3
goroutine_1:释放
锁
goroutine_2：获取
锁
goroutine_2:set
timeseat[1]=4