CODEWITHSUNDEEP
go
curiousengineer
curiousengineer

Reputation: 2607

Go channel not deadlocked or locked

I have a small snippet of code like this

func main() {
    var c chan string
    go func() {
        c <- "let's get started"//write1
        //c <- "let's get started"//write2
        //c <- "let's get started"//write3
        fmt.Println("wrote the stuff....")
    }()
    //time.Sleep(3 * time.Second) //adding this always shows fatal exception
    c = make(chan string)
    fmt.Println(<-c)
}

I do not see the output wrote the stuff.... on console if I uncomment //write2 and write3 code snippet lines. I understand that probably I don't see it is because the channel is unbuffered and totally synchronous, and there is only one read happening outside the channel. However, when program exits, the go routine is blocked, why is there no error like deadlocked... or something in this case that I see?

Upvotes: 3

Views: 89

Answers (2)

Siddharth Shishulkar
Siddharth Shishulkar

Reputation: 131

This is an interesting experiment. I don't know if its possible but this type of error should be caught by the compiler, there should be a feature that checks at the compile-time if anything is writing in a nil channel.

Upvotes: -2

user2357112
user2357112

Reputation: 280778

Create the channel before you try to write to it. If the func goroutine tries to send elements through c before there's actually a channel there, it ends up using a nil channel, which blocks forever.

Upvotes: 6

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