在Go中使用互斥锁 [英] Using mutexes in Go

问题描述

我试图了解互斥的工作原理.到目前为止，据我了解，它是用来进行原子操作并同步访问某些数据的.

I am trying to understand how mutexes work. From my understanding so far, it is made to carry atomic operations and synchronize access to some data.

我在此处构建了队列数据结构的示例: https://github.com/arnauddri/algorithms/blob/master/data-structures%2Fqueue%2Fqueue.go

I built an example of a queue data structure here: https://github.com/arnauddri/algorithms/blob/master/data-structures%2Fqueue%2Fqueue.go

下面是一些代码:

package queue

import "sync"

type Queue struct {
    queue []interface{}
    len   int
    lock  *sync.Mutex
}

func New() *Queue {
    queue := &Queue{}
    queue.queue = make([]interface{}, 0)
    queue.len = 0

    return queue
}

func (q *Queue) Push(el interface{}) {
    q.lock.Lock()
    defer q.lock.Unlock()

    q.queue = append(q.queue, el)
    q.len++
}

但是，当我尝试创建队列并将项目推送到其中时，出现运行时错误:

However when I try to create a queue and push an item to it I get a runtime error:

q := New()
q.Push(1)

panic: runtime error: invalid memory address or nil pointer dereference [recovered]
panic: runtime error: invalid memory address or nil pointer dereference

我真的不明白这里发生了什么.

I really don't understand what is happening here.

在这里我应该如何使用Mutex?

How should I use the Mutex here?

非常感谢

推荐答案

问题似乎在于您从未实例化互斥体.当您运行 New()函数时，您正在创建一个空队列，其中包含一个可以引用互斥锁的变量，但是您实际上并没有告诉它这样做，这意味着此时queue.lock == nil .您可以通过在 New()函数中添加实例化行来解决此问题.

Looks like the issue is that you're never instantiating the mutex. When you run the New() function you're creating an empty Queue with a variable that can reference a mutex, but you never actually tell it to do so, which means that at this point queue.lock == nil. You can fix this by adding in an instantiation line to your New() function.

queue.lock = new(sync.Mutex)

这是一个有效的游乐场演示: http://play.golang.org/p/Qa6buDaHIj

Here is a playground demo that works: http://play.golang.org/p/Qa6buDaHIj

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