Why does the condition contained in the mutex part not deadlock in this golang example?

WBOY
Release: 2024-02-13 18:06:10
forward
833 people have browsed it

为什么在这个 golang 示例中,互斥体部分中包含的条件不会出现死锁?

In this golang example, the reason why deadlock will not occur with the conditions contained in the mutex part is because the mutex is passed through `Lock()` and `Unlock() ` method to achieve mutually exclusive access to shared resources. When a goroutine calls the `Lock()` method, if the mutex is already locked by another goroutine, the goroutine will be blocked until the mutex is released. This blocking mechanism ensures that when the mutex is locked, multiple goroutines will not access shared resources at the same time, thereby avoiding the occurrence of deadlock. So in this example, due to the correct usage of mutex, the condition will not be deadlocked.

Question content

I saw this example during a training at O'Reilly. There is a condition that should prevent widgetInventory from becoming negative. The example works, but I don't understand why the program doesn't deadlock when makeSales gets the mutex and widgetInventory is 0.

var (
    wg sync.WaitGroup
    mutex = sync.Mutex{}
    widgetInventory int32= 1000
    newPurchase = sync.NewCond(&mutex)
)

func main() {
    fmt.Println("Starting inventory count = ", widgetInventory)
    wg.Add(2)
    go makeSales()
    go newPurchases()
    wg.Wait()
    fmt.Println("Ending inventory count = ", widgetInventory)
}

func makeSales() {
    for i := 0; i < 3000; i++ {
        mutex.Lock()
        if widgetInventory-100 < 0{
            newPurchase.Wait()
        }
        widgetInventory -= 100
        fmt.Println(widgetInventory)
        mutex.Unlock()
    }
    wg.Done()
}

func newPurchases() {
    for i := 0; i < 3000; i++ {
        mutex.Lock()
        widgetInventory+= 100
        fmt.Println(widgetInventory)
        newPurchase.Signal()
        mutex.Unlock()
    }
    wg.Done()
}
Copy after login

I expect the code to deadlock when makeSales gets the mutex and widgetInventory is 0.

Solution

I didn't notice that the condition was associated with the mutex: newPurchase =sync.NewCond(&mutex) Entering .Wait() unlocks the mutex and attempts to reacquire it when the condition signal is received.

condition.Wait() can only be used when acquiring a mutex, so it works at the cost of less readable code :-)

The above is the detailed content of Why does the condition contained in the mutex part not deadlock in this golang example?. For more information, please follow other related articles on the PHP Chinese website!

source:stackoverflow.com
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template
About us Disclaimer Sitemap
php.cn:Public welfare online PHP training,Help PHP learners grow quickly!