例行泄漏会发生在具有一个缓冲器的通道中，该缓冲器具有两个输入但只有一个输出吗？

Question

I have a function that is used to forward a message between two io.ReadWriters. Once an error happens, I need to log the error and return. But I think I may have a goroutine leakage problem in my code:

func transport(rw1, rw2 io.ReadWriter) error {
    errc := make(chan error, 1) // only one buffer
    go func() {
        _, err := io.Copy(rw1, rw2)
        errc <- err
    }()

    go func() {
        _, err := io.Copy(rw2, rw1)
        errc <- err
    }()

    err := <-errc // only one error catched
    if err != nil && err == io.EOF {
        err = nil
    }
    return err
}

Because there is only one error can be caught in this function, will the second goroutine exit and garbaged normally? Or should I write one more err <- errc to receive another error.

Answer 1

The value from one goroutine is received and the other is buffered. Both goroutines can send to the channel and exit. There is no leak.

You might want to receive both values to ensure that application detects an error when the first goroutine to send is successful and the second goroutine encounters an error.

var err error
for i := 0; i < 2; i++ {
  if e := <-errc; e != nil {
    err = e
  }
}

Because io.Copy does not return io.EOF, there's no need to check for io.EOF when collecting the errors.

The code can be simplified to use a single goroutine:

errc := make(chan error, 1)
go func() {
    _, err := io.Copy(rw1, rw2)
    errc <- err
}()

_, err := io.Copy(rw2, rw1)

if e := <-errc; e != nil {
   err = e
}