Channel Synchronization, waitgroups

How can you ensure that the following program only exists after the go-routine has executed -

package main

func expensiveOp() {
  sum := 0
  for i := range 1000000 {
    sum += i
  }
  println(sum)
}

func main() {
  go expensiveOp()
  println("main Function ended")
}

Solution - Add an unbuffered channel that blocks the receiver

package main

func expensiveOp(ch chan<- bool) {
  sum := 0
  for i := range 1000000 {
    sum += i
  }
  println(sum)
  ch <- true
}

func main() {
  ch := make(chan bool)
  go expensiveOp(ch)
  <-ch
  println("main Function ended")
}

Waitgroups

In Go, a WaitGroup is a synchronization primitive that allows you to wait for a collection of goroutines to finish executing.

Write a program that runs 5 different go routines and waits for all of them to exit before printing “done”

Ugly approach (without waitgroups)

package main

func expensiveOp(ch chan<- bool) {
  sum := 0
  for i := range 1000000 {
    sum += i
  }
  println(sum)
  ch <- true
}

func main() {
  ch1 := make(chan bool)
  go expensiveOp(ch1)

  ch2 := make(chan bool)
  go expensiveOp(ch2)

  ch3 := make(chan bool)
  go expensiveOp(ch3)

  ch4 := make(chan bool)
  go expensiveOp(ch4)

  ch5 := make(chan bool)
  go expensiveOp(ch5)

  <-ch1
  <-ch2
  <-ch3
  <-ch4
  <-ch5
  println("main Function ended")
}

Better approach (Using waitgroups)

package main

import "sync"

func expensiveOp() {
  sum := 0
  for i := range 1000000 {
    sum += i
  }
  println(sum)
}

func main() {
  var wg sync.WaitGroup
  for i := 0; i < 5; i++ {
    wg.Add(1)

    go func() {
      defer wg.Done()
      expensiveOp()
    }()
  }
  wg.Wait()
  println("main Function ended")
}

Assignment

Add a timeout to a go-routine so it gets ignored and execution continues if it takes more than 10 seconds