Race condition with waitgroup and unbuffered channel - go

Question

After getting (the right) solution to my initial problem in this post Understanding golang channels: deadlock, I have come up with a slightly different solution (which in my opinion reads better:
// Binary histogram counts the occurences of each word.
package main
import (
"fmt"
"strings"
"sync"
)
var data = []string{
"The yellow fish swims slowly in the water",
"The brown dog barks loudly after a drink ...",
"The dark bird bird of prey lands on a small ...",
}
func main() {
histogram := make(map[string]int)
words := make(chan string)
var wg sync.WaitGroup
for _, line := range data {
wg.Add(1)
go func(l string) {
for _, w := range strings.Split(l, " ") {
words <- w
}
wg.Done()
}(line)
}
go func() {
for w := range words {
histogram[w]++
}
}()
wg.Wait()
close(words)
fmt.Println(histogram)
}
It does work, but unfortunately running it against race, it shows 2 race conditions:
==================
WARNING: DATA RACE
Read at 0x00c420082180 by main goroutine:
...
Previous write at 0x00c420082180 by goroutine 9:
...
Goroutine 9 (running) created at:
main.main()
Can you help me understand where is the race condition?

Answer 1

You are trying to read from histogram in fmt.Println(histogram) which is not synchronized to the write of the goroutine mutating it histogram[w]++. You can add a lock to synchronize the writes and reads.
e.g.
var lock sync.Mutex
go func() {
lock.Lock()
defer lock.Unlock()
for w := range words {
histogram[w]++
}
}()
//...
lock.Lock()
fmt.Println(histogram)
Note you can also use a sync.RWMutex.
Another thing you could do is to wait for the goroutine mutating histogram to finish.
var histWG sync.WaitGroup
histWG.Add(1)
go func() {
for w := range words {
histogram[w]++
}
histWG.Done()
}()
wg.Wait()
close(words)
histWG.Wait()
fmt.Println(histogram)
Or simply use a channel to wait.
done := make(chan bool)
go func() {
for w := range words {
histogram[w]++
}
done <- true
}()
wg.Wait()
close(words)
<-done
fmt.Println(histogram)