[日常] Go语言圣经-竞争条件习题

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
}