Go语言圣经-竞争条件习题详解编程语言

package main 
import( 
	"fmt" 
	"sync" 
) 
var balance int 
func Deposit(amount int) { balance = balance + amount } 
func Balance() int { return balance } 
/* 
问题: 
1.在Alice运行期间 balance = balance + amount 这一步运算可能会被Bob中间挤占 
2.当运行到balance + amount的时候,Bob的正好赶到,然后继续运行blance= 
3.此时Bob的增加的数据会丢失 
*/ 
func main(){ 
	var wg sync.WaitGroup 
	wg.Add(1) 
	// Alice: 
	go func() { 
		defer wg.Done() 
	    Deposit(200)                // A1 
	    fmt.Println("=", Balance()) // A2 
	}() 
	wg.Add(1) 
	// Bob: 
	go func(){ 
		defer wg.Done() 
		Deposit(100) 
	}() 
	wg.Wait() 
	res:=Balance() 
	fmt.Println(res) 
} 

package main 
import( 
	"fmt" 
	"sync" 
) 
var balance int 
 
var deposits = make(chan int) //存款用channel 
var balances = make(chan int) //接收余额用channel 
 
 
func Deposit(amount int) {deposits <- amount} 
func Balance() int { return <-balances } 
 
func main(){ 
	go teller() 
 
	var wg sync.WaitGroup 
	wg.Add(1) 
	go func(){ 
		defer wg.Done() 
		Deposit(100) 
		fmt.Println("=",Balance()) 
	}() 
	wg.Add(1) 
	go func(){ 
		defer wg.Done() 
		Deposit(200) 
		fmt.Println("=",Balance()) 
	}() 
	wg.Add(1) 
	go func(){ 
		defer wg.Done() 
		res:=Withdraw(200) 
		if !res{ 
			fmt.Println("取款失败") 
		} 
	}() 
	wg.Wait() 
	b:=Balance() 
	fmt.Println(b) 
} 
/* 
解决: 
1.总余额限定在一个goroutine中,通过channel通讯 
2.channel是会阻塞同一时间的多个goroutine的 
*/ 
func teller() { 
    var balance int //总余额限定在一个goroutine中 
    for { 
        select { 
        case amount := <-deposits: 
            balance += amount 
        case balances <- balance: 
        } 
    } 
} 
//取款用函数 
func Withdraw(amount int)bool{ 
    Deposit(-amount) 
    if Balance() < 0 { 
        Deposit(amount) 
        return false // insufficient funds 
    } 
    return true 
}