面试题：Go语言中互斥锁死锁预防之场景分析

Question

在Go语言并发编程中，假设有两个goroutine，goroutine A持有互斥锁`mutexA`，并试图获取互斥锁`mutexB`；goroutine B持有互斥锁`mutexB`，并试图获取互斥锁`mutexA`，这种情况下可能会发生死锁。请编写一段代码来模拟这个场景，并阐述如何通过调整获取锁的顺序来预防死锁。

Answer 1

模拟死锁代码

package main

import (
    "fmt"
    "sync"
)

var (
    mutexA sync.Mutex
    mutexB sync.Mutex
)

func goroutineA(wg *sync.WaitGroup) {
    defer wg.Done()
    mutexA.Lock()
    fmt.Println("goroutine A acquired mutexA")
    mutexB.Lock()
    fmt.Println("goroutine A acquired mutexB")
    mutexB.Unlock()
    mutexA.Unlock()
}

func goroutineB(wg *sync.WaitGroup) {
    defer wg.Done()
    mutexB.Lock()
    fmt.Println("goroutine B acquired mutexB")
    mutexA.Lock()
    fmt.Println("goroutine B acquired mutexA")
    mutexA.Unlock()
    mutexB.Unlock()
}

func main() {
    var wg sync.WaitGroup
    wg.Add(2)
    go goroutineA(&wg)
    go goroutineB(&wg)
    wg.Wait()
}

预防死锁方案

通过调整获取锁的顺序来预防死锁，比如让两个goroutine都按照相同的顺序获取锁。

调整后代码

package main

import (
    "fmt"
    "sync"
)

var (
    mutexA sync.Mutex
    mutexB sync.Mutex
)

func goroutineA(wg *sync.WaitGroup) {
    defer wg.Done()
    mutexA.Lock()
    fmt.Println("goroutine A acquired mutexA")
    mutexB.Lock()
    fmt.Println("goroutine A acquired mutexB")
    mutexB.Unlock()
    mutexA.Unlock()
}

func goroutineB(wg *sync.WaitGroup) {
    defer wg.Done()
    mutexA.Lock()
    fmt.Println("goroutine B acquired mutexA")
    mutexB.Lock()
    fmt.Println("goroutine B acquired mutexB")
    mutexB.Unlock()
    mutexA.Unlock()
}

func main() {
    var wg sync.WaitGroup
    wg.Add(2)
    go goroutineA(&wg)
    go goroutineB(&wg)
    wg.Wait()
}

在上述修改后的代码中，goroutineA和goroutineB都先获取mutexA，再获取mutexB，这样就避免了死锁的发生。