This recursive function works as expected (returns 5 lines with numbers 5 to 1):
package main
import (
"fmt"
)
func recur(iter int) {
if iter <= 0 {
return
}
fmt.Println(iter)
recur(iter-1)
}
func main() {
recur(5)
}
this one does not (returns only 1 line with number 5):
package main
import (
"fmt"
)
func recur(iter int) {
if iter <= 0 {
return
}
fmt.Println(iter)
go recur(iter-1)
}
func main() {
recur(5)
}
The difference is that in the second implementation, function calls itself as a goroutine. (line go recur(iter-1)
)
So can someone explain this behaviour?
If you make everything asynchronous, there is nothing in main
to wait on. You have to explicitly wait on the go routines so your program does not exit before the recursive process finishes.
Use sync.WaitGroup
or something similar for synchronization. Example (On Play):
func recur(iter int, g *sync.WaitGroup) {
defer g.Done()
if iter <= 0 {
return
}
fmt.Println(iter)
go recur(iter-1, g)
}
func main() {
g := &sync.WaitGroup{}
runs := 5
g.Add(runs)
recur(runs, g)
g.Wait()
}
Your programming is exiting before all the goroutines finish. The simplest way to see this work is as follows.
package main
import (
"fmt"
"time"
)
func recur(iter int) {
if iter <= 0 {
return
}
fmt.Println(iter)
go recur(iter - 1)
}
func main() {
recur(5)
time.Sleep(time.Second)
}
playground link
Rather than sleeping, however, you may want to pass a sync.WaitGroup
through to your function.
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