Minimal readline with timeout example

readln.go

import (
	"bufio"
	"errors"
	"strings"
	"time"
)

// Readln reads and returns a single line (sentinal: \n) from stdin.
// If a given timeout has passed, an error is returned.
// This functionality is similar to GNU's `read -e -p [s] -t [num] [name]`
func Readln(prompt string, timeout time.Duration) ([]byte, error) {
	s := make(chan string)
	e := make(chan error)

	go func() {
		fmt.Print(prompt)
		reader := bufio.NewReader(os.Stdin)
		line, err := reader.ReadString('\n')
		if err != nil {
			e <- err
		} else {
			s <- line
		}
		close(s)
		close(e)
	}()

	select {
	case line := <-s:
		return line, nil
	case err := <-e:
		return nil, err
		case <-time.After(timeout):
		return nil, errors.New("Timeout")
	}
}

You can also use bufio.NewScanner() and grab sc.Text().

reader.ReadString() seems never exit in this case