https://play.golang.org/p/5A-dnZVy2aA
func closeChannel(stream chan int) {
fmt.Println("closing the channel")
close(stream)
}
func main() {
chanOwner := func() <-chan int {
resultStream := make(chan int, 5)
go func() {
defer closeChannel(resultStream)
for i := 0; i <= 5; i++ {
resultStream <- i
}
}()
return resultStream
}
resultStream := chanOwner()
for result := range resultStream { //this blocks until the channel is closed
fmt.Printf("Received: %d\n", result)
}
fmt.Println("done receiving")
}
Output of program
closing the channel
Received: 0
Received: 1
Received: 2
Received: 3
Received: 4
Received: 5
done receiving
why is the closing the channel statement printed before any Received in the above program. since the channel is buffered with a capacity of 5, and we are inserting 6 elements into it, I was expecting it to block at the resultStream <- i before a value was read and cleared up space in the buffer.
The generator goroutine fills the channel to its capacity and blocks. The receiver for loops receives the first item from the channel, which enables the generator goroutine again. The generator goroutine runs to completion before the receiver for-loop can print the Received message, printing the closing the channel message. Then the receiver loop receives all remaining messages.
Related
If I pass more than 12 for totalValues to getResults(), no values are read from channel resultCh and I get fatal error: all goroutines are asleep - deadlock!:
package main
type result struct {
index int
count int
}
func getResults(indexCh chan int, resultCh chan result, totalValues int) {
allFields := make([]int, totalValues)
for i := range allFields {
indexCh <- i
println("Sent value", i)
}
println("Sent all values")
for i := 0; i < totalValues; i++ {
value := <-resultCh
println("Received result", value.index)
allFields[value.index] = value.count
}
println("Done processing")
}
func processValue(indexCh chan int, ch chan result) {
for index := range indexCh {
println("Received value", index)
ch <- result{
index: index,
count: index * index,
}
println("Sent result", index)
}
}
func main() {
bufferDepth := 4
indexCh := make(chan int, bufferDepth)
resultCh := make(chan result, bufferDepth)
for i := 0; i < 4; i++ {
go processValue(indexCh, resultCh)
}
// Value > 12 results in goroutine asleep
getResults(indexCh, resultCh, 12)
}
The above code works by launching four goroutines using function processValue() that read values from buffered channel indexCh. Integers 0 to totalValues are inserted into indexCh by getResults(). processValue() processes the value and puts the result onto buffered channel resultCh, which is read by a getResults().
I only only observe this problem when totalValues is greater than 12. No values are read from resultCh because "Received result..." is not printed.
If I increase bufferDepth to five, the program completes successfully for totalValues > 12 and < 15.
The program also completes successfully if the buffer depth of resultCh matches totalValues.
getResults starts by writing all values to the channel. There are 4 goroutines listening, so each picks up those values, and they are blocked writing to the result channel. Channels have buffer sizes of 4. So 4 goroutines each can write a total of 4 values to result channel before blocking.
4 goroutines + 4 index channel size + 4 result channel size = 12 entries
After that, the processValue is blocked at writing to the channel, because getResults are also blocked writing to the indexCh.
I am working on a sample program to print sum of odd and sum of even number between 1 to 100 using goroutine with multiple channels.
you can find my code
here
output
sum of even number = 2550
sum of odd number = 2500
fatal error: all goroutines are asleep - deadlock!
goroutine 1 [chan receive]:
main.print(0x434100, 0x11db7c)
/tmp/sandbox052575152/main.go:18 +0xc0
main.main()
/tmp/sandbox052575152/main.go:14 +0x120
The code works but with deadlock.
I am not sure what is wrong in my code
We can iterate through values sent over a channel. To break such iteration channel needs to be closed explicitly. Otherwise range would block forever in the same way as for nil channel. In your code you did't close the sum(for print function sumValues channel) channel. That's why following function will be blocked for forever.
func print(sumValues <-chan string ){
for val := range sumValues {
fmt.Println(val)
}
}
So you have to close the sum channel in the doSum function after all the go routine in the doSum function are complete (otherwise sum channel might be closed before go routines are complete). You can use sync.WaitGroup to do that. See the updated doSum function below:
func doSum(sum chan<- string, oddChan <-chan int, evenChan <-chan int) {
var waitGroup sync.WaitGroup
waitGroup.Add(2) // Must wait for 2 calls to 'done' before moving on
go func(sum chan<- string) {
s1 := 0
for val := range oddChan {
s1 += val
}
sum <- fmt.Sprint("sum of odd number = ", s1)
waitGroup.Done()
}(sum)
go func(sum chan<- string) {
s1 := 0
for val := range evenChan {
s1 += val
}
sum <- fmt.Sprint("sum of even number = ", s1)
waitGroup.Done()
}(sum)
// Waiting for all goroutines to exit
waitGroup.Wait()
// all goroutines are complete now close the sum channel
close(sum)
}
I thought that channels in Go only hold 1 value by default unless the buffer size if specified. I read that here. But when I run this:
func main (){
for i := range numGen(6) {
log.Println("taking from channel", i)
}
}
func numGen(num int) chan int {
c := make(chan string)
go func() {
for i := 0; i < num; i++ {
log.Println("passing to channel", i)
c <- i
}
close(c)
}
return c
}
my output is:
2017/06/13 18:09:08 passing to channel 0
2017/06/13 18:09:08 passing to channel 1
2017/06/13 18:09:08 taking from channel 0
2017/06/13 18:09:08 taking from channel 1
2017/06/13 18:09:08 passing to channel 2
2017/06/13 18:09:08 passing to channel 3
2017/06/13 18:09:08 taking from channel 2
2017/06/13 18:09:08 taking from channel 3
2017/06/13 18:09:08 passing to channel 4
2017/06/13 18:09:08 passing to channel 5
2017/06/13 18:09:08 taking from channel 4
2017/06/13 18:09:08 taking from channel 5
which shows that the channel is holding 2 values at a time. Specifying a buffer size like this
c := make(chan int, 0)
does nothing. Any way I could make it only hold 1, value, not 2?
which shows that the channel is holding 2 values at a time.
Thats not the case. This is how the code executes:
The main goroutine blocks on a read on the channel
The second goroutine writes to the channel and continues executing.
The second goroutine blocks at the second write attempt because noone is reading
The main goroutine continues executing, prints the read number
The main goroutine reads another number since someone is writing to it
The main goroutine prints the read number and blocks on the next read
The second goroutine continues executing at step 2.
There is no buffer, just concurrency.
package main
import (
"log"
)
func main() {
seq := make(chan bool)
for i := range numGen(6, seq) {
<-seq
log.Println("taking from channel", i)
}
}
func numGen(num int, seq chan bool) chan int {
c := make(chan int)
go func() {
for i := 0; i < num; i++ {
c <- i
log.Println("passing to channel", i)
seq <- true // 要保证顺序，这里发送一个信号量。
}
close(c)
}()
return c
}
I'm reading/working through Go Concurrency Patterns: Pipelines and cancellation, but i'm having trouble understanding the Stopping short section. We have the following functions:
func sq(in <-chan int) <-chan int {
out := make(chan int)
go func() {
for n := range in {
out <- n * n
}
close(out)
}()
return out
}
func gen(nums ...int) <-chan int {
out := make(chan int)
go func() {
for _, n := range nums {
out <- n
}
close(out)
}()
return out
}
func merge(cs ...<-chan int) <-chan int {
var wg sync.WaitGroup
out := make(chan int, 1) // enough space for the unread inputs
// Start an output goroutine for each input channel in cs. output
// copies values from c to out until c is closed, then calls wg.Done.
output := func(c <-chan int) {
for n := range c {
out <- n
}
wg.Done()
}
wg.Add(len(cs))
for _, c := range cs {
go output(c)
}
// Start a goroutine to close out once all the output goroutines are
// done. This must start after the wg.Add call.
go func() {
wg.Wait()
close(out)
}()
return out
}
func main() {
in := gen(2, 3)
// Distribute the sq work across two goroutines that both read from in.
c1 := sq(in)
c2 := sq(in)
// Consume the first value from output.
out := merge(c1, c2)
fmt.Println(<-out) // 4 or 9
return
// Apparently if we had not set the merge out buffer size to 1
// then we would have a hanging go routine.
}
Now, if you notice line 2 in merge, it says we make the out chan with buffer size 1, because this is enough space for the unread inputs. However, I'm almost positive that we should allocate a chan with buffer size 2. In accordance with this code sample:
c := make(chan int, 2) // buffer size 2
c <- 1 // succeeds immediately
c <- 2 // succeeds immediately
c <- 3 // blocks until another goroutine does <-c and receives 1
Because this section implies that a chan of buffer size 3 would not block. Can anyone please clarify/assist my understanding?
The program sends two values to the channel out and reads one value from the channel out. One of the values is not received.
If the channel is unbuffered (capacity 0), then one of the sending goroutines will block until the program exits. This is a leak.
If the channel is created with a capacity of 1, then both goroutines can send to the channel and exit. The first value sent to the channel is received by main. The second value remains in the channel.
If the main function does not receive a value from the channel out, then a channel of capacity 2 is required to prevent the goroutines from blocking indefinitely.
I am trying to understand, how golang channel works. I read a book about the go language and found the following example.
package main
import (
"fmt"
)
// Send the sequence 2, 3, 4, ... to returned channel
func generate() chan int {
ch := make(chan int)
go func() {
for i := 2; i <= 100 ; i++ {
ch <- i
}
}()
return ch
}
// Filter out input values divisible by 'prime', send rest to returned channel
func filter(in chan int, prime int) chan int {
out := make(chan int)
go func() {
for {
if i := <-in; i%prime != 0 {
out <- i
}
}
}()
return out
}
func sieve() chan int {
out := make(chan int)
go func() {
ch := generate()
for {
prime := <-ch
ch = filter(ch, prime)
out <- prime
}
}()
return out
}
func main() {
primes := sieve()
for {
fmt.Println(<-primes)
}
}
When I run this programm, I've got a deadlock, but when I change the generate function to
// Send the sequence 2, 3, 4, ... to returned channel
func generate() chan int {
ch := make(chan int)
go func() {
for i := 2; ; i++ {
ch <- i
}
}()
return ch
}
Then the programm will run the infinite loop, but not deadlock. Why do I get deadlock, when I remove the condition in for loop?
What do you mean with blocking principle?
You can see it illustrated in the blog post "The Nature Of Channels In Go "
for an unbuffered channel:
(Illustration from blog post "The Nature Of Channels In Go ", written by William Kennedy, Feb. 2014)
Unbuffered channels have no capacity and therefore require both goroutines to be ready to make any exchange.
When a goroutine attempts to write a resource to an unbuffered channel and there is no goroutine waiting to receive the resource, the channel will lock the goroutine and make it wait.
When a goroutine attempts to read from an unbuffered channel, and there is no goroutine waiting to send a resource, the channel will lock the goroutine and make it wait.
That is what happens in your case with your reader:
func main() {
primes := sieve()
for {
fmt.Println(<-primes)
}
}
since primes is never closed, main remains blocked.
It (main) is in step 3:
in step 3, the goroutine on the right places his hand into the channel or performs a read.
That goroutine is also locked in the channel until the exchange is complete.
The sender never calls close(primes).
Let's consider a simpler example:
func generate() chan int {
ch := make(chan int)
go func() {
for i := 2; /*i < 100*/; i++ {
ch <- i
}
}()
return ch
}
func main() {
for i := range generate() {
fmt.Println(i)
}
}
With the condition i < 100 uncommented, the goroutine spawned by generate stops after sending 98 numbers. However, it does not close the channel, so main has no way of knowing that no more numbers are going to be sent, and it just keeps blocking on the channel. Since main is now the only goroutine still in existence (the other one has returned), and it's blocking, you have a deadlock.