How can I use channel send direction in Go

Question

In Go, it can be specified which direction a channel can send. I am trying to create an example about it, look at the following code:

package main

import (
    "fmt"
    "time"
)

func main() {

    ic_send_only := make(<-chan int) //a channel that can only send data - arrow going out is sending
    ic_recv_only := make(chan<- int) //a channel that can only receive a data - arrow going in is receiving

    go func() {
        ic_recv_only <- 4555
    }()

    go func() {

        ic_send_only <- ic_recv_only

    }()

    fmt.Println(ic_recv_only)
    time.Sleep(10000)

}

I get the compiler error

# command-line-arguments
.\send_receive.go:19: invalid operation: ic_send_only <- ic_recv_only (send to receive-only type <-chan int)
[Finished in 0.2s with exit code 2]

How can I use channel direction in the right way?

Or does anyone have a better sample than me?

Answer 1

Three issues:

• You have the send and receive operations reversed (which is the error you're seeing)
• Creating recv-only or send-only channels make no sense, as you cannot use them

• The notation you're using is trying to send the channel itself, not the result. You need to receive and send, which requires two arrows.

  ic_recv_only <- <-ic_send_only

You may be confused because you have the terminology reversed. <-ch is a "receive operation", and ch <- is a send operation. Note that in your example, everything would be deadlocked because you can't complete the corresponding sends and receives to pass something through either channel.

Here is a complete example:

// This receives an int from a channel. The channel is receive-only
func consumer(ch <-chan int) int {
    return <-ch
}

// This sends an int over a channel. The channel is send-only
func producer(i int, ch chan<- int) {
    ch <- i
}

func main() {
    ch := make(chan int)
    go producer(42, ch)
    result := consumer(ch)
    fmt.Println("received", result)
}