I am trying to make a terminal golang application, where a user has 4 seconds to input something. If he inputted something faster, print result and ask him input again for 4 seconds.
If a user will not return input in 4 seconds, the program must write time out and ask him input again.
My code does this, but only once. After the first timeout it won't return any result even if a user was faster that 4 seconds. I cannot figure out why this is so.
The code
package main
import (
"bufio"
"fmt"
"log"
"os"
"time"
)
var (
result string
err error
)
func getInput(input chan string) {
in := bufio.NewReader(os.Stdin)
result, err := in.ReadString('
')
if err != nil {
log.Fatal(err)
}
input <- result
}
func main() {
for {
fmt.Println("input something")
input := make(chan string, 1)
go getInput(input)
select {
case i := <-input:
fmt.Println("result")
fmt.Println(i)
case <-time.After(4000 * time.Millisecond):
fmt.Println("timed out")
}
}
}
The output:
input something
123
result
123
input something
2
result
2
input something
timed out
input something
2
timed out
input something
timed out
input something
The problem has to do with the way you're getting the user's input. On a timeout you spawn a new go routine asking for input, but the old one that you had spawned previously is still there grabbing input and sending it to a channel that no one is listening to any more.
Changing it to something like this would fix the problem:
func getInput(input chan string) {
for {
in := bufio.NewReader(os.Stdin)
result, err := in.ReadString('
')
if err != nil {
log.Fatal(err)
}
input <- result
}
}
func main() {
input := make(chan string, 1)
go getInput(input)
for {
fmt.Println("input something")
select {
case i := <-input:
fmt.Println("result")
fmt.Println(i)
case <-time.After(4000 * time.Millisecond):
fmt.Println("timed out")
}
}
}