How to break out of select gracefuly in golang - go

Question

I have a program in golang that counts SHA1s and prints ones that start with two zeros. I want to use goroutines and channels. My problem is that I don't know how to gracefully exit select clause if I don't know how many results it will produce.
Many tutorials know that in advance and exit when counter hits. Other suggest using WaitGroups, but I don't want to do that: I want to print results in main thread as soon it appears in channel. Some suggest to close a channel when goroutines are finished, but I want to close it after asynchronous for finishes, so I don't know how.
Please help me to achieve my requirements:
package main
import (
"crypto/sha1"
"fmt"
"time"
"runtime"
"math/rand"
)
type Hash struct {
message string
hash [sha1.Size]byte
}
var counter int = 0
var max int = 100000
var channel = make(chan Hash)
var source = rand.NewSource(time.Now().UnixNano())
var generator = rand.New(source)
func main() {
nCPU := runtime.NumCPU()
runtime.GOMAXPROCS(nCPU)
fmt.Println("Number of CPUs: ", nCPU)
start := time.Now()
for i := 0 ; i < max ; i++ {
go func(j int) {
count(j)
}(i)
}
// close channel here? I can't because asynchronous producers work now
for {
select {
// how to stop receiving if there are no producers left?
case hash := <- channel:
fmt.Printf("Hash is %v\n ", hash)
}
}
fmt.Printf("Count of %v sha1 took %v\n", max, time.Since(start))
}
func count(i int) {
random := fmt.Sprintf("This is a test %v", generator.Int())
hash := sha1.Sum([]byte(random))
if (hash[0] == 0 && hash[1] == 0) {
channel <- Hash{random, hash}
}
}

Answer 1

Firstly: if you don't know when your computation ends, how could you even model it? Make sure you know exactly when and under what circumstances your program terminates. If you're done you know how to write it in code.
You're basically dealing with a producer-consumer problem. A standard case. I would model
that this way (on play):
Producer
func producer(max int, out chan<- Hash, wg *sync.WaitGroup) {
defer wg.Done()
for i := 0; i < max; i++ {
random := fmt.Sprintf("This is a test %v", rand.Int())
hash := sha1.Sum([]byte(random))
if hash[0] == 0 && hash[1] == 0 {
out <- Hash{random, hash}
}
}
close(out)
}
Obviously you're brute-forcing hashes, so the end is reached when the loop is finished.
We can close the channel here and signal the other goroutines that there is nothing more to listen for.
Consumer
func consumer(max int, in <-chan Hash, wg *sync.WaitGroup) {
defer wg.Done()
for {
hash, ok := <-in
if !ok {
break
}
fmt.Printf("Hash is %v\n ", hash)
}
}
The consumer takes all the incoming messages from the in channel and checks if it was closed (ok).
If it is closed, we're done. Otherwise print the received hashes.
Main
To start this all up we can write:
wg := &sync.WaitGroup{}
c := make(chan Hash)
wg.Add(1)
go producer(max, c, wg)
wg.Add(1)
go consumer(max, c, wg)
wg.Wait()
The WaitGroup's purpose is to wait until the spawned goroutines finished, signalled by
the call of wg.Done in the goroutines.
Sidenote
Also note that the Rand you're using is not safe for concurrent access. Use the one initialized
globally in math/rand. Example:
rand.Seed(time.Now().UnixNano())
rand.Int()

Answer 2

The structure of your program should probably be re-examined.
Here is a working example of what I presume you are looking for.
It can be run on the Go playground
package main
import (
"crypto/sha1"
"fmt"
"math/rand"
"runtime"
"time"
)
type Hash struct {
message string
hash [sha1.Size]byte
}
const Max int = 100000
func main() {
nCPU := runtime.NumCPU()
runtime.GOMAXPROCS(nCPU)
fmt.Println("Number of CPUs: ", nCPU)
hashes := Generate()
start := time.Now()
for hash := range hashes {
fmt.Printf("Hash is %v\n ", hash)
}
fmt.Printf("Count of %v sha1 took %v\n", Max, time.Since(start))
}
func Generate() <-chan Hash {
c := make(chan Hash, 1)
go func() {
defer close(c)
source := rand.NewSource(time.Now().UnixNano())
generator := rand.New(source)
for i := 0; i < Max; i++ {
random := fmt.Sprintf("This is a test %v", generator.Int())
hash := sha1.Sum([]byte(random))
if hash[0] == 0 && hash[1] == 0 {
c <- Hash{random, hash}
}
}
}()
return c
}
Edit: This does not fire up a separate routine for each Hash computation,
but to be honest, I fail to see the value on doing so. The scheduling of all those routines will likely cost you far more than running the code in a single routine.
If need be, you can split it up into chunks of N routines, but a 1:1 mapping is not the way to go with this.