Welcome to tutorial no. 23 in Golang tutorial series.

What are buffered channels?

All the channels we discussed in the previous tutorial were basically unbuffered. As we discussed in the channels tutorial in detail, sends and receives to an unbuffered channel are blocking.

It is possible to create a channel with a buffer. Sends to a buffered channel are blocked only when the buffer is full. Similarly receives from a buffered channel are blocked only when the buffer is empty.

Buffered channels can be created by passing an additional capacity parameter to the make function which specifies the size of the buffer.

ch := make(chan type, capacity)

capacity in the above syntax should be greater than 0 for a channel to have a buffer. The capacity for an unbuffered channel is 0 by default and hence we omitted the capacity parameter while creating channels in the previous tutorial.

Lets write some code and create a buffered channel.

Example

package main import ( "fmt" ) func main() { ch := make(chan string, 2) ch <- "naveen" ch <- "paul" fmt.Println(<- ch) fmt.Println(<- ch) }

Run program in playground

In the program above, in line no. 9 we create a buffered channel with a capacity of 2. Since the channel has a capacity of 2, it is possible to write 2 strings into the channel without being blocked. We write 2 strings to the channel in line no. 10 and 11 and the channel does not block. We read the 2 strings written in line nos. 12 and 13 respectively. This program prints,

naveen paul

Another Example

Lets look at one more example of buffered channel in which the values to the channel are written in a concurrent Goroutine and read from the main Goroutine. This example will help us better understand when writes to a buffered channel block.

package main import ( "fmt" "time" ) func write(ch chan int) { for i := 0; i < 5; i++ { ch <- i fmt.Println("successfully wrote", i, "to ch") } close(ch) } func main() { ch := make(chan int, 2) go write(ch) time.Sleep(2 * time.Second) for v := range ch { fmt.Println("read value", v,"from ch") time.Sleep(2 * time.Second) } }

Run program in playground





In the program above, a buffered channel ch of capacity 2 is created in line no. 16 of the main Goroutine and passed to the write Goroutine in line no. 17. Then the main Goroutine sleeps for 2 seconds. During this time, the write Goroutine is running concurrently. The write Goroutine has a for loop which writes numbers from 0 to 4 to the ch channel. The capacity of this buffered channel is 2 and hence the write Goroutine will be able to write values 0 and 1 to the ch channel immediately and then it blocks until at least one value is read from ch channel. So this program will print the following 2 lines immediately.

successfully wrote 0 to ch successfully wrote 1 to ch

After printing the above two lines, the writes to the ch channel in the write Goroutine are blocked until someone reads from the ch channel. Since the main Goroutine sleeps for 2 seconds before starting to read from the channel, the program will not print anything for the next 2 seconds. The main Goroutine wakes up after 2 seconds and starts reading from the ch channel using a for range loop in line no. 19, prints the read value and then sleeps for 2 seconds again and this cycle continues until the ch is closed. So the program will print the following lines after 2 seconds,

read value 0 from ch successfully wrote 2 to ch

This will continue until all values are written to the channel and it is closed in the write Goroutine. The final output would be,

successfully wrote 0 to ch successfully wrote 1 to ch read value 0 from ch successfully wrote 2 to ch read value 1 from ch successfully wrote 3 to ch read value 2 from ch successfully wrote 4 to ch read value 3 from ch read value 4 from ch

Deadlock

package main import ( "fmt" ) func main() { ch := make(chan string, 2) ch <- "naveen" ch <- "paul" ch <- "steve" fmt.Println(<-ch) fmt.Println(<-ch) }

Run program in playground

In the program above, we write 3 strings to a buffered channel of capacity 2. When the control reaches the third write in line no. 11, the write is blocked since the channel has exceeded its capacity. Now some Goroutine must read from the channel in order for the write to proceed, but in this case there is no concurrent routine reading from this channel. Hence there will be a deadlock and the program will panic at run time with the following message,

fatal error: all goroutines are asleep - deadlock! goroutine 1 [chan send]: main.main() /tmp/sandbox274756028/main.go:11 +0x100

Length vs Capacity

The capacity of a buffered channel is the number of values that the channel can hold. This is the value we specify when creating the buffered channel using the make function.

The length of the buffered channel is the number of elements currently queued in it.

A program will make things clear 😀

package main import ( "fmt" ) func main() { ch := make(chan string, 3) ch <- "naveen" ch <- "paul" fmt.Println("capacity is", cap(ch)) fmt.Println("length is", len(ch)) fmt.Println("read value", <-ch) fmt.Println("new length is", len(ch)) }

Run program in playground

In the program above, the channel is created with a capacity of 3 , that is, it can hold 3 strings. We then write 2 strings to the channel in line nos. 9 and 10 respectively. Now the channel has 2 strings queued in it and hence its length is 2 . In line no. 13, we read a string from the channel. Now the channel has only one string queued in it and hence its length becomes 1 . This program will print,

capacity is 3 length is 2 read value naveen new length is 1

WaitGroup

The next section in this tutorial is about Worker Pools. To understand worker pools, we need to first know about WaitGroup as it will be used in the implementation of Worker pool.

A WaitGroup is used to wait for a collection of Goroutines to finish executing. The control is blocked until all Goroutines finish executing. Lets say we have 3 concurrently executing Goroutines spawned from the main Goroutine. The main Goroutines needs to wait for the 3 other Goroutines to finish before terminating. This can be accomplished using WaitGroup.





Lets stop the theory and write some code right away 😀

package main import ( "fmt" "sync" "time" ) func process(i int, wg *sync.WaitGroup) { fmt.Println("started Goroutine ", i) time.Sleep(2 * time.Second) fmt.Printf("Goroutine %d ended

", i) wg.Done() } func main() { no := 3 var wg sync.WaitGroup for i := 0; i < no; i++ { wg.Add(1) go process(i, &wg) } wg.Wait() fmt.Println("All go routines finished executing") }

Run in playground

WaitGroup is a struct type and we are creating a zero value variable of type WaitGroup in line no.18. The way WaitGroup works is by using a counter. When we call Add on the WaitGroup and pass it an int , the WaitGroup 's counter is incremented by the value passed to Add . The way to decrement the counter is by calling Done() method on the WaitGroup. The Wait() method blocks the Goroutine in which it's called until the counter becomes zero.

In the above program, we call wg.Add(1) in line no. 20 inside the for loop which iterates 3 times. So the counter now becomes 3. The for loop also spawns 3 process Goroutines and then wg.Wait() called in line no. 23 makes the main Goroutine to wait until the counter becomes zero. The counter is decremented by the call to wg.Done in the process Goroutine in line no. 13. Once all the 3 spawned Goroutines finish their execution, that is once wg.Done() has been called three times, the counter will become zero, and the main Goroutine will be unblocked.

It is important to pass the address of wg in line no. 21. If the address is not passed, then each Goroutine will have its own copy of the WaitGroup and main will not be notified when they finish executing.

This program outputs.

started Goroutine 2 started Goroutine 0 started Goroutine 1 Goroutine 0 ended Goroutine 2 ended Goroutine 1 ended All go routines finished executing

Your output might be different from mine since the order of execution of Goroutines can vary :).

Worker Pool Implementation

One of the important uses of buffered channel is the implementation of worker pool.

In general, a worker pool is a collection of threads which are waiting for tasks to be assigned to them. Once they finish the task assigned, they make themselves available again for the next task.

We will implement worker pool using buffered channels. Our worker pool will carry out the task of finding the sum of a digits of the input number. For example if 234 is passed, the output would be 9 (2 + 3 + 4). The input to the worker pool will be list of pseudo random integers.

The following are the core functionalities of our worker pool

Creation of a pool of Goroutines which listen on an input buffered channel waiting for jobs to be assigned

Addition of jobs to the input buffered channel

Writing results to an output buffered channel after job completion

Read and print results from the output buffered channel

We will write this program step by step to make it easier to understand.

The first step will be creation of the structs representing the job and the result.

type Job struct { id int randomno int } type Result struct { job Job sumofdigits int }

Each Job struct has a id and a randomno for which the sum of the individual digits has to be computed.

The Result struct has a job field which is the job for which it holds the result (sum of individual digits) in the sumofdigits field.

The next step is to create the buffered channels for receiving the jobs and writing the output.

var jobs = make(chan Job, 10) var results = make(chan Result, 10)

Worker Goroutines listen for new tasks on the jobs buffered channel. Once a task is complete, the result is written to the results buffered channel.

The digits function below does the actual job of finding the sum of the individual digits of an integer and returning it. We will add a sleep of 2 seconds to this function just to simulate the fact that it takes some time for this function to calculate the result.

func digits(number int) int { sum := 0 no := number for no != 0 { digit := no % 10 sum += digit no /= 10 } time.Sleep(2 * time.Second) return sum }

Next we will write a function which creates a worker Goroutine.

func worker(wg *sync.WaitGroup) { for job := range jobs { output := Result{job, digits(job.randomno)} results <- output } wg.Done() }

The above function creates a worker which reads from the jobs channel, creates a Result struct using the current job and the return value of the digits function and then writes the result to the results buffered channel. This function takes a WaitGroup wg as parameter on which it will call the Done() method when all jobs have been completed.

The createWorkerPool function will create a pool of worker Goroutines.

func createWorkerPool(noOfWorkers int) { var wg sync.WaitGroup for i := 0; i < noOfWorkers; i++ { wg.Add(1) go worker(&wg) } wg.Wait() close(results) }

The function above takes the number of workers to be created as a parameter. It calls wg.Add(1) before creating the Goroutine to increment the WaitGroup counter. Then it creates the worker Goroutines by passing the address of the WaitGroup wg to the worker function. After creating the needed worker Goroutines, it waits for all the Goroutines to finish their execution by calling wg.Wait() . After all Goroutines finish executing, it closes the results channel since all Goroutines have finished their execution and no one else will further be writing to the results channel.

Now that we have the worker pool ready, lets go ahead and write the function which will allocate jobs to the workers.

func allocate(noOfJobs int) { for i := 0; i < noOfJobs; i++ { randomno := rand.Intn(999) job := Job{i, randomno} jobs <- job } close(jobs) }

The allocate function above takes the number of jobs to be created as input parameter, generates pseudo random numbers with a maximum value of 998 , creates Job struct using the random number and the for loop counter i as the id and then writes them to the jobs channel. It closes the jobs channel after writing all jobs.

Next step would be to create the function that reads the results channel and prints the output.

func result(done chan bool) { for result := range results { fmt.Printf("Job id %d, input random no %d , sum of digits %d

", result.job.id, result.job.randomno, result.sumofdigits) } done <- true }

The result function reads the results channel and prints the job id, input random no and the sum of digits of the random no. The result function also takes a done channel as parameter to which it writes to once it has printed all the results.

We have everything set now. Lets go ahead and finish the last step of calling all these functions from the main() function.

func main() { startTime := time.Now() noOfJobs := 100 go allocate(noOfJobs) done := make(chan bool) go result(done) noOfWorkers := 10 createWorkerPool(noOfWorkers) <-done endTime := time.Now() diff := endTime.Sub(startTime) fmt.Println("total time taken ", diff.Seconds(), "seconds") }

We first store the execution start time of the program in line no.2 of the main function and in the last line (line no. 12) we calculate the time difference between the endTime and startTime and display the total time it took for the program to run. This is needed because we will do some benchmarks by changing the number of Goroutines.

The noOfJobs is set to 100 and then allocate is called to add jobs to the jobs channel.

Then done channel is created and passed to the result Goroutine so that it can start printing the output and notify once everything has been printed.

Finally a pool of 10 worker Goroutines are created by the call to createWorkerPool function and then main waits on the done channel for all the results to be printed.

Here is the full program for your reference. I have imported the necessary packages too.

package main import ( "fmt" "math/rand" "sync" "time" ) type Job struct { id int randomno int } type Result struct { job Job sumofdigits int } var jobs = make(chan Job, 10) var results = make(chan Result, 10) func digits(number int) int { sum := 0 no := number for no != 0 { digit := no % 10 sum += digit no /= 10 } time.Sleep(2 * time.Second) return sum } func worker(wg *sync.WaitGroup) { for job := range jobs { output := Result{job, digits(job.randomno)} results <- output } wg.Done() } func createWorkerPool(noOfWorkers int) { var wg sync.WaitGroup for i := 0; i < noOfWorkers; i++ { wg.Add(1) go worker(&wg) } wg.Wait() close(results) } func allocate(noOfJobs int) { for i := 0; i < noOfJobs; i++ { randomno := rand.Intn(999) job := Job{i, randomno} jobs <- job } close(jobs) } func result(done chan bool) { for result := range results { fmt.Printf("Job id %d, input random no %d , sum of digits %d

", result.job.id, result.job.randomno, result.sumofdigits) } done <- true } func main() { startTime := time.Now() noOfJobs := 100 go allocate(noOfJobs) done := make(chan bool) go result(done) noOfWorkers := 10 createWorkerPool(noOfWorkers) <-done endTime := time.Now() diff := endTime.Sub(startTime) fmt.Println("total time taken ", diff.Seconds(), "seconds") }

Run in playground

Please run this program in your local machine for more accuracy in the total time taken calculation.

This program will print,

Job id 1, input random no 636, sum of digits 15 Job id 0, input random no 878, sum of digits 23 Job id 9, input random no 150, sum of digits 6 ... total time taken 20.01081009 seconds

A total of 100 lines will be printed corresponding to the 100 jobs and then finally the total time taken for the program to run will be printed in the last line. Your output will differ from mine as the Goroutines can run in any order and the total time will also vary based on the hardware. In my case it takes approximately 20 seconds for the program to complete.

Now lets increase the noOfWorkers in the main function to 20 . We have doubled the number the workers. Since the worker Goroutines have increased(doubled to be precise), the total time taken for the program to complete should reduce(by half to be precise). In my case it became, 10.004364685 seconds and the program printed,

... total time taken 10.004364685 seconds

Now we can understand that as the number of worker Goroutines increase, the total time taken to complete the jobs decreases. I leave it as an exercise for you to play with the noOfJobs and noOfWorkers in the main function to different values and analyse the results.

This brings us to an end of this tutorial. Have a good day.

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