Reputation: 4352
Playing around with Go's channels and routines I have come across a peculiar behaviour I was hoping somebody could explain.
Below is a short program that is supposed to print a couple strings to stdout, by sending the strings through a channel to a "listener" (the select statement) running in a separate goroutine.
package main
import (
"fmt"
"time"
)
func main() {
a := make(chan string)
go func() {
for {
select {
case <-a:
fmt.Print(<-a)
}
}
}()
a <- "Hello1\n"
a <- "Hello2\n"
a <- "Hello3\n"
a <- "Hello4\n"
time.Sleep(time.Second)
}
Using
go func() {
for s := range a {
fmt.Print(s)
}
}()
// or even simpler
go func() {
for {
fmt.Print(<-a)
}
}()
works as expected. However, running the uppermost snippet with the select statement produces the following output:
Hello2
Hello4
i.e. only every other statement is printed. What kind of sorcery is this?
Upvotes: 6
Views: 396
Reputation: 11
package main
import (
"fmt"
"time"
)
func main() {
a := make(chan string)
go func() {
for {
select {
case v:= <-a:
fmt.Print(v)
}
}
}()
a <- "Hello1\n"
a <- "Hello2\n"
a <- "Hello3\n"
a <- "Hello4\n"
time.Sleep(5*time.Second)
}
Upvotes: 0
Reputation: 18917
<-a
gets a value from the channel, destructively. So in your code you get two values, one in the select statement, and one to print. The one received in the select statement is not bound to any variable, and is therefore lost.
Try
select {
case val := <-a:
fmt.Print(val)
instead, to get only one value, bind it to variable val, and print it out.
Upvotes: 7
Reputation: 38781
In the uppermost snippet, you're pulling two values from the channel for each loop. One in the select statement and one in the print statement.
Change
select {
case <-a:
fmt.Print(<-a)
To
select {
case val := <-a:
fmt.Print(val)
http://play.golang.org/p/KIADcwkoKs
Upvotes: 12