sync.Mutex如何在并发Golang程序中工作

Question

When working with a concurrent program in Go, we use var mutex sync.Mutex and then if we have to write execute some synchronized code block, we call mutex.Lock() and mutex.Unlock(). Now I have some doubts here:

1) Is there only one instance of mutex i.e. if mutex is being used for locking for operation of CodeBlockA, and at the same time some other goroutine B has to use it for some other CodeBlockB that also uses mutex, will goroutine B be blocked till goroutine A relieves mutex.

2) Is it possible that we can have more instances of muteness guarding their own dedicated code blocks.

Answer 1

While writing the sample code as reference, I got the answer. if we have var mu sync.Mutex then Locking and Unlocking of mu will be serialized. It cannot be locked at the same time at two places which could be totally independent code bases. but if we have var mu1 sync.Mutex and var mu2 sync.Mutex then mu1 and mu2 can be locked concurrently. Following is a small program for the POC:

package main

import (
    "fmt"
    "sync"
    "time"
)

var mu sync.Mutex
var mu1 sync.Mutex
var wg sync.WaitGroup

// var anotherBalance int

func code1() {
    mu.Lock()
    defer mu.Unlock()
    fmt.Println("Entering code1")
    time.Sleep(1 * time.Second)
    fmt.Println("Exiting code1")
}

func code2() {
    mu1.Lock()
    defer mu1.Unlock()
    fmt.Println("Entering code2")
    time.Sleep(1 * time.Second)
    fmt.Println("Exiting code2")
}

func main() {
    wg.Add(1)
    go func() {
        code1()
        wg.Done()
    }()
    wg.Add(1)
    go func() {
        code2()
        wg.Done()
    }()
    wg.Wait()
}

And after running it,

Entering code2

Entering code1

Exiting code2

Exiting code1

real 0m1.335s

user 0m0.307s

sys 0m0.069s