[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)
}

练习 9.1：给gopl.io/ch9/bank1程序添加一个Withdraw(amount int)取款函数。其返回结果应该要表明事务是成功了还是因为没有足够资金失败了。这条消息会被发送给monitor的goroutine，且消息需要包含取款的额度和一个新的channel，这个新channel会被monitor goroutine来把boolean结果发回给Withdraw。

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
}