CODEWITHSUNDEEP
go
puoklam
puoklam

Reputation: 25

Goroutine buffered channel deadlock

In the main goroutine, I'm getting a deadlock if I send values to channel exceeded the capacity, but if I'm sending in a different goroutine, there will be no deadlock, Why?

func main() {
    c := make(chan int, 2)
    for i := 0; i < 3; i++ {
        c <- i
    }
}

func main() {
    c := make(chan int, 2)
    go func() {
        for i := 0; i < 3; i++ {
            c <- i
        }
        close(c)
    }()
    time.Sleep(5 * time.Second)
}

Upvotes: 0

Views: 63

Answers (1)

Barys Malaichyk
Barys Malaichyk

Reputation: 164

func main() {
   c := make(chan int, 2)
   for i := 0; i < 3; i++ {
       // Blocking operation
       c <- i
   }
}

By default >>>sends<<< and receives block(!) until both the sender and receiver are ready

Checkout - https://gobyexample.com/channels

There are no receivers on your channel, main goroutine(= sender) is blocked

Upvotes: 0

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