Channel synchronization in Go - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to send a value into the channel.

Go

ch := make(chan int)

func main() {
    go func() {
        ch[1] 42
    }()
    val := <-ch
    println(val)
}

Drag options to blanks, or click blank then click option'

A<-chan

B<-

C<-ch

D<-chan int

Attempts:

3 left

💡 Hint

Common Mistakes

Using the receive operator '<-' instead of the send operator.

Confusing channel type syntax with send/receive syntax.

To send a value into a channel in Go, use the syntax ch <- value.

2fill in blank

medium

Complete the code to receive a value from the channel.

Go

ch := make(chan string)

func main() {
    go func() {
        ch <- "hello"
    }()
    msg := [1]ch
    println(msg)
}

Drag options to blanks, or click blank then click option'

A<-ch

B<-

Cch

Dchan

Attempts:

3 left

💡 Hint

Common Mistakes

Using the channel name alone without the receive operator.

Using the send operator instead of receive operator.

To receive a value from a channel, use the syntax value := <-ch.

3fill in blank

hard

Fix the error in the code to properly synchronize using a channel.

Go

done := make(chan bool)

func main() {
    go func() {
        println("Working...")
        [1] <- true
    }()
    <-done
}

Drag options to blanks, or click blank then click option'

Adone

Bdone()

Cdone <-

D<-done

Attempts:

3 left

💡 Hint

Common Mistakes

Reversing the send operator direction.

Calling the channel as a function.

To send a value into the channel, use done <- true. The channel variable name must be on the left.

4fill in blank

hard

Fill both blanks to create a map of word lengths for words longer than 3 characters.

Go

words := []string{"go", "code", "chat", "ai"}
lengths := map[string]int{
    [1]: [2] for _, word := range words if len(word) > 3
}

Drag options to blanks, or click blank then click option'

A"word"

Bword

Clen(word)

Dword.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Using string literals as keys instead of variables.

Using incorrect method syntax like word.len().

The key should be the word itself, and the value should be its length using len(word).

5fill in blank

hard

Fill all three blanks to create a synchronized counter using a channel.

Go

counter := 0
increment := make(chan bool)

go func() {
    for {
        <-[1]
        [2]++
        println([3])
    }
}()

func main() {
    increment <- true
}

Drag options to blanks, or click blank then click option'

Aincrement

Bcounter

Dincrement()

Attempts:

3 left

💡 Hint

Common Mistakes

Using function call syntax for the channel.

Mixing up variable names for counter and channel.

The channel to receive from is increment. The counter variable is incremented and printed.

Practice

1. What happens when a goroutine tries to send data on an unbuffered channel if no other goroutine is ready to receive?

easy

A. The sending goroutine waits until another goroutine receives the data.

B. The data is lost and the program continues immediately.

C. The program crashes with a runtime error.

D. The data is buffered automatically for later use.

Solution

1. Step 1: Understand unbuffered channel behavior

   Unbuffered channels require both sender and receiver to be ready at the same time for communication.

2. Step 2: Analyze sending on unbuffered channel

   If no goroutine is ready to receive, the sending goroutine blocks and waits until a receiver is ready.

3. Final Answer:

   The sending goroutine waits until another goroutine receives the data. -> Option A

4. Quick Check:

   Unbuffered channel send blocks until receive [OK]

Hint: Unbuffered sends block until a receiver is ready [OK]

Common Mistakes:

• Thinking data is buffered automatically
• Assuming send never blocks
• Believing program crashes on send

2. Which of the following is the correct syntax to create a buffered channel of integers with capacity 3 in Go?

easy

A. ch := chan int{3}

B. ch := make(chan int)

C. ch := make(chan int, 3)

D. ch := new(chan int, 3)

Solution

1. Step 1: Recall channel creation syntax

   Channels are created using make with type and optional buffer size: make(chan Type, capacity).

2. Step 2: Identify correct buffered channel syntax

   To create a buffered channel of int with capacity 3, use make(chan int, 3).

3. Final Answer:

   ch := make(chan int, 3) -> Option C

4. Quick Check:

   Buffered channel syntax = make(chan Type, capacity) [OK]

Hint: Use make(chan Type, capacity) for buffered channels [OK]

Common Mistakes:

• Omitting buffer size for buffered channel
• Using new() instead of make()
• Incorrect channel literal syntax

3. What will be the output of the following Go program?

package main
import "fmt"
func main() {
  ch := make(chan string)
  go func() {
    ch <- "hello"
  }()
  msg := <-ch
  fmt.Println(msg)
}

medium

A. Compilation error

B. No output

C. Deadlock runtime error

D. hello

Solution

1. Step 1: Analyze channel and goroutine

   A goroutine sends "hello" on an unbuffered channel ch, while main receives from ch.

2. Step 2: Understand synchronization

   The send and receive synchronize, so main receives "hello" and prints it.

3. Final Answer:

   hello -> Option D

4. Quick Check:

   Send and receive synchronize, printing message [OK]

Hint: Unbuffered channel send/receive synchronize to pass data [OK]

Common Mistakes:

• Expecting deadlock because of goroutine
• Thinking channel is buffered by default
• Assuming no output due to concurrency

4. Identify the error in this Go code snippet involving channel synchronization:

package main
func main() {
  ch := make(chan int)
  ch <- 5
  val := <-ch
  println(val)
}

medium

A. No error, code runs fine.

B. Sending on channel blocks forever causing deadlock.

C. Channel is not initialized properly.

D. Receiving from channel blocks forever causing deadlock.

Solution

1. Step 1: Check channel creation and usage

   Channel ch is unbuffered and created correctly with make(chan int).

2. Step 2: Analyze send operation

   Sending on unbuffered channel blocks until another goroutine receives. Here, send happens before receive, no other goroutine receives, so it blocks forever.

3. Final Answer:

   Sending on channel blocks forever causing deadlock. -> Option B

4. Quick Check:

   Send blocks without concurrent receiver [OK]

Hint: Send on unbuffered channel blocks if no receiver ready [OK]

Common Mistakes:

• Thinking receive blocks first
• Assuming channel is nil
• Believing code runs without deadlock

5. You want to run three tasks concurrently and wait for all to finish before exiting. Which Go code snippet correctly uses channels to synchronize all tasks?

hard

A. Use a buffered channel with capacity 3; each goroutine sends a signal when done; main receives 3 signals.

B. Use an unbuffered channel; main sends signals to goroutines to start; no synchronization on finish.

C. Use a buffered channel with capacity 1; goroutines send signals; main receives once.

D. Use no channels; just start goroutines and exit main immediately.

Solution

1. Step 1: Understand synchronization goal

   Main must wait for all 3 goroutines to finish before exiting.

2. Step 2: Analyze channel usage options

   Buffered channel with capacity 3 allows all goroutines to send done signals without blocking; main receives 3 signals to know all finished.

3. Final Answer:

   Use a buffered channel with capacity 3; each goroutine sends a signal when done; main receives 3 signals. -> Option A

4. Quick Check:

   Buffered channel capacity = number of goroutines to sync [OK]

Hint: Use buffered channel sized to goroutines for sync signals [OK]

Common Mistakes:

• Using unbuffered channel causing deadlock
• Receiving fewer signals than goroutines
• Not waiting for goroutines at all