5

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?

RSH
  • 504
  • 5
  • 9

2 Answers2

11

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()
}
nemo
  • 55,207
  • 13
  • 135
  • 135
  • 3
    When traversing a graph you have an unknown amount of branches in each recursion. Possible refactor: call `Add(1)` before each `recur()`, including main. See example: https://play.golang.org/p/OuZVgowAbt – Eran W Oct 27 '17 at 21:12
1

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.

Amit Kumar Gupta
  • 17,184
  • 7
  • 46
  • 64