golang并发编程实践 -- 简单生产者消费者(with lock)

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上一篇文章用golang中的channel实现了简单的消费者模型,下面的版本是用传统的锁技术实现的版本,相对比会发现golang提供的channel更好用。而且golang的channel可以完成很多在别的语言里需要很多代码才能实现的功能。以后陆续解答。

package main

import (
	"fmt"
	"sync"
	"time"
)

type Queue struct {
	Elem     []int
	Capacity int
	Front    int
	Rear     int
	Lock     sync.Locker
	Cond     *sync.Cond
}

func New() *Queue {
	theQueue := &Queue{}
	theQueue.Capacity = 10
	theQueue.Elem = make([]int,10)
	theQueue.Front,theQueue.Rear = 0,0
	theQueue.Lock = &sync.Mutex{}
	theQueue.Cond = sync.NewCond(theQueue.Lock)
	return theQueue
}

func (self *Queue) Put(e int) {
	self.Cond.L.Lock()
	// the Queue is full,Producer waits here
	// note that we use for not if to test the condition
	for self.Full() {
		self.Cond.Wait()
	}

	self.Elem[self.Rear] = e
	self.Rear = (self.Rear + 1) % self.Capacity
	self.Cond.Signal()
	defer self.Cond.L.Unlock()
}

func (self *Queue) Get() int {
	self.Cond.L.Lock()
	// the Queue is empty,Consumer waits here
	// note that we use for not if to test the condition
	for self.Empty() {
		self.Cond.Wait()
	}

	p := self.Elem[self.Front]
	self.Front = (self.Front + 1) % self.Capacity
	self.Cond.Signal()
	defer self.Cond.L.Unlock()
	return p
}

func (self *Queue) Empty() bool {
	if self.Front == self.Rear {
		return true
	}
	return false
}

func (self *Queue) Full() bool {
	if ((self.Rear + 1) % self.Capacity) == self.Front {
		return true
	}
	return false
}

func main() {
	theQueue := New()

	// producer puts
	go func() {
		for i := 1; i <= 100; i++ {
			time.Sleep(100 * time.Millisecond)
			theQueue.Put(i)
			fmt.Println("Bob puts ",i)
		}
	}()

	// consumer gets
	for i := 1; i <= 100; i++ {
		time.Sleep(100 * time.Millisecond)
		p := theQueue.Get()
		fmt.Println("Alice gets : ",p)
	}
}

运行效果如下:
Bob puts  1
Alice gets :  1
Bob puts  2
Alice gets :  2
Bob puts  3
Alice gets :  3
Bob puts  4
Alice gets :  4
Bob puts  5
Alice gets :  5
Bob puts  6
Alice gets :  6
Bob puts  7
Alice gets :  7
Bob puts  8
Alice gets :  8
Bob puts  9
Alice gets :  9
Bob puts  10
Alice gets :  10
Bob puts  11
Alice gets :  11
Bob puts  12
Alice gets :  12
Bob puts  13
Alice gets :  13

.......

如此反复直到100次。

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