defer、goto以及sync.Mutex

前言

最近写着写着遇到些稀奇古怪的问题，想着与defer、GOTO以及锁有关，于是整理了一下。

切入正题

首先来看看本文的主角们

十分好用的defer

func main() {
    doSomething()
    defer finishDoSomething()   
    println("Something has been done")
}

func doSomething() {
    println("Do something")
}

func finishDoSomething() {
    println("Finish do something")
}

这是一个defer的用例，很简单，就是在doSomething结束之后，再执行finishDOsomething（println输出到stderr,fmt.Println输出到stdout,我绝对不是为了偷懒少打点才用的println）,以上代码的输出如下：

Do something
Something has been done
Finish do somethin

没有任何问题，十分正常且符合预期的输出。

掺和一下goto，顺便加了点料

现在笔者想在doSomething结束后，从头来过，再来一遍，为了便于区分，顺便加了点其他料：

func main() {
    restart := true
    runningTimes := 1
INIT:
    doSomething(runningTimes)
    defer finishDoSomething(runningTimes)
    println("Something has been done ", runningTimes, " time")
    if restart {
        println("Star Over")
        restart = false
        runningTimes++
        goto INIT
    }
}

func doSomething(time int) {
    println("Do something ", time, " time")
}

func finishDoSomething(time int) {
    println("Finish do something ", time, " time")
}

此时程序的输出为

Do something  1  time
Something has been done  1  time
Star Over
Do something  2  time
Something has been done  2  time
Finish do something  2  time
Finish do something  1  time

前五行看下来似乎都没什么问题，着重看最后两行，为什么runningTimes变成2之后仍旧打出了
Finish do something 1 time呢?
这里其实与goto没有任何关系，是defer的特性，当程序经过defer时，会进行参数预计算，也就是说，当runTimes是1时，加载到defer finishDoSomething(runningTimes)的时候，就会把finishDoSomething(1)丢到栈里，所以最后可以打印出来Finish do something 1 time。
然后是打印次序及顺序的问题，把刚刚的执行过程用代码重新梳理一下可能就会好理解地多：

doSomething(1)
defer finishDoSomething(1)
println("Something has been done ", 1, " time")
println("Star Over")
doSomething(2)
defer finishDoSomething(2)
println("Something has been done ", 2, " time")

所以，goto 之前有defer的照样会执行，很显然，在这个例子里执行了两次defer（当然预计算的值不一样）。

此时一个人畜无害的Mutex混入其中

此时，出于某种原因，在doSomething的时候，必须对locker进行锁定，好让其他人不能同时一起doSomething，稍加修改：

func main() {
	// 不想多写一个goroutine绝对不是我懒
    restart := true
    runningTimes := 1
    locker := sync.Mutex{}
INIT:
    locker.Lock()
    doSomething(runningTimes)
    defer func() {
        finishDoSomething(runningTimes)
        defer locker.Unlock()
    }()
    println("Something has been done ", runningTimes, " time")
    if restart {
        println("Star Over")
        restart = false
        runningTimes++
        goto INIT
    }
}

func doSomething(time int) {
    println("Do something ", time, " time")
}

func finishDoSomething(time int) {
    println("Finish do something ", time, " time")
}

此时执行此程序，似乎出现了不得了的问题(经典故弄玄虚环节)

Do something  1  time
Something has been done  1  time
Star Over
fatal error: all goroutines are asleep - deadlock!

竟然出现了死锁！(似乎也不必这么惊讶)
这里面的死锁其实是由于锁重入导致的，Golang是不支持重入锁的，这就会导致死锁。
已知了原因似乎也很好解决了，大不了goto前解锁呗：

func main() {
	// 不想多写一个goroutine绝对不是我懒
    restart := true
    runningTimes := 1
    locker := sync.Mutex{}
INIT:
    locker.Lock()
    doSomething(runningTimes)
    defer func() {
        finishDoSomething(runningTimes)
        defer locker.Unlock()
    }()
    println("Something has been done ", runningTimes, " time")
    if restart {
        println("Star Over")
        restart = false
        runningTimes++
        locker.Unlock()
        goto INIT
    }
}

func doSomething(time int) {
    println("Do something ", time, " time")
}

func finishDoSomething(time int) {
    println("Finish do something ", time, " time")
}

于是梅开二度：

Do something  1  time
Something has been done  1  time
Star Over
Do something  2  time
Something has been done  2  time
Finish do something  2  time
Finish do something  2  time
fatal error: sync: unlock of unlocked mutex

确实，解决了重入问题，但是没有解决Unlock两次的问题。
从程序跑通的角度来看，利用defer的特性再加一个defer locker.Lock()便可解君愁了：

func main() {
	// 不想多写一个goroutine绝对不是我懒
    restart := true
    runningTimes := 1
    locker := sync.Mutex{}
INIT:
    locker.Lock()
    doSomething(runningTimes)
    defer func() {
        finishDoSomething(runningTimes)
        defer locker.Unlock()
    }()
    println("Something has been done ", runningTimes, " time")
    if restart {
        println("Star Over")
        restart = false
        runningTimes++
        locker.Unlock()
        defer locker.Lock()
        goto INIT
    }
}

func doSomething(time int) {
    println("Do something ", time, " time")
}

func finishDoSomething(time int) {
    println("Finish do something ", time, " time")
}

成功了！它跑起来了！开心！(棒读)
原本加locker的目的在于执行本段代码的时候，有且只有一个goroutine在执行，但是在上述的修改中，更加积重难返，为了Unlock特意Lock这种操作其实大可不必，到这种程度了，修改原来代码的逻辑其实是更好的选择，比如：

func main(){
    restart := true
    runningTimes := 1
    locker := sync.Mutex{}
    locker.Lock()
    defer locker.Unlock()
INIT:
    doSomething(runningTimes)
    defer func() {
        finishDoSomething(runningTimes)
    }()
    println("Something has been done ", runningTimes, " time")
    if restart {
        println("Star Over")
        restart = false
        runningTimes++
        goto INIT
    }    
}

又或者您坚持需要一个可以重入的锁建议自己写一个，可以参考一下以下解决方案：
Go语言如何实现可重入锁？ - SegmentFault 思否