Blocking behavior in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Go program with channels?

Consider the following Go code. What will it print when run?

Go

package main
import (
	"fmt"
)
func main() {
	ch := make(chan int)
	go func() {
		ch <- 42
	}()
	val := <-ch
	fmt.Println(val)
}

A 42

B 0

C Compilation error

D Deadlock panic

💡 Hint

Think about how channels block until data is sent or received.

The goroutine sends 42 into the channel. The main goroutine waits to receive from the channel, so it blocks until the value is sent. Then it prints 42.

❓ Predict Output

intermediate

What happens when reading from a closed channel?

What will this Go program print?

Go

package main
import "fmt"
func main() {
	ch := make(chan int)
	close(ch)
	val, ok := <-ch
	fmt.Println(val, ok)
}

A 0 true

B 0 false

C Compilation error

D Deadlock panic

💡 Hint

Reading from a closed channel returns zero value and false.

When a channel is closed, receiving from it returns the zero value of the channel type and false for the second value.

❓ Predict Output

advanced

What is the output of this select with blocking channels?

Analyze the following Go code and determine its output.

Go

package main
import (
	"fmt"
	"time"
)
func main() {
	ch1 := make(chan string)
	ch2 := make(chan string)
	go func() {
		time.Sleep(100 * time.Millisecond)
		ch1 <- "first"
	}()
	go func() {
		time.Sleep(50 * time.Millisecond)
		ch2 <- "second"
	}()
	select {
	case msg1 := <-ch1:
		fmt.Println(msg1)
	case msg2 := <-ch2:
		fmt.Println(msg2)
	}
}

A first

B Compilation error

C second

D Deadlock panic

💡 Hint

Which channel sends first after the sleep?

ch2 sends after 50ms, ch1 after 100ms. The select receives from the first ready channel, so it prints "second".

❓ Predict Output

advanced

What error does this Go program produce?

What happens when you run this Go code?

Go

package main
func main() {
	ch := make(chan int)
	ch <- 1
}

A Deadlock panic

B Compilation error

C Program prints 1

D No output, program exits normally

💡 Hint

Sending on an unbuffered channel blocks until another goroutine receives.

The main goroutine tries to send on ch but no other goroutine receives, so it blocks forever and causes a deadlock panic.

🧠 Conceptual

expert

How many items are in the buffered channel after this code runs?

Given the following Go code, how many items remain in the channel after execution?

Go

package main
import "fmt"
func main() {
	ch := make(chan int, 3)
	ch <- 10
	ch <- 20
	<-ch
	ch <- 30
	fmt.Println(len(ch))
}

A 3

B 0

C 1

D 2

💡 Hint

Remember that reading from the channel removes an item.

Two items remain because one was removed by the receive operation, so the channel holds 2 items.

Practice

1. What does blocking mean in Go programming?

easy

A. Running multiple tasks at the same time

B. Waiting for an operation to finish before moving on

C. Ignoring errors in the program

D. Skipping a function call

Solution

1. Step 1: Understand the meaning of blocking

   Blocking means the program waits for something to finish before continuing.

2. Step 2: Match the definition to options

   Waiting for an operation to finish before moving on correctly describes waiting for an operation to finish before moving on.

3. Final Answer:

   Waiting for an operation to finish before moving on -> Option B

4. Quick Check:

   Blocking = Waiting [OK]

Hint: Blocking means waiting, not running or skipping [OK]

Common Mistakes:

• Confusing blocking with concurrency
• Thinking blocking means ignoring errors
• Assuming blocking skips code

2. Which of the following Go code snippets correctly shows a blocking receive on a channel?

easy

A. value := <-ch

B. go func() { <-ch }()

C. ch <- value

D. select { case ch <- value: }

Solution

1. Step 1: Identify blocking receive syntax

   Receiving from a channel blocks until a value is available, done by value := <-ch.

2. Step 2: Check other options

   go func() { <-ch }() launches a goroutine, so receive is not blocking main code. ch <- value sends, not receives. select { case ch <- value: } sends in select, not receive.

3. Final Answer:

   value := <-ch -> Option A

4. Quick Check:

   Blocking receive = value := <-ch [OK]

Hint: Blocking receive uses '<-ch' without goroutine [OK]

Common Mistakes:

• Confusing send and receive syntax
• Thinking goroutine makes receive blocking
• Using select for receive incorrectly

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

package main
import "fmt"
func main() {
  ch := make(chan int)
  go func() {
    ch <- 42
  }()
  fmt.Println(<-ch)
}

medium

A. 42

B. 0

C. Compilation error

D. Deadlock runtime error

Solution

1. Step 1: Analyze channel creation and goroutine

   A channel ch is created and a goroutine sends 42 into it.

2. Step 2: Understand blocking receive

   Main goroutine waits on <-ch until value 42 is sent, then prints it.

3. Final Answer:

   42 -> Option A

4. Quick Check:

   Send and receive synchronize, output = 42 [OK]

Hint: Send in goroutine, receive blocks until value arrives [OK]

Common Mistakes:

• Expecting deadlock without goroutine
• Thinking channel defaults to zero value
• Confusing send and receive order

4. Identify the problem in this Go code and fix it:

package main
func main() {
  ch := make(chan int)
  ch <- 10
  val := <-ch
  println(val)
}

medium

A. Compilation error due to missing import; fix by adding fmt package

B. Runtime panic due to nil channel; fix by initializing channel

C. No problem; code runs and prints 10

D. Deadlock because send blocks without receiver; fix by using a goroutine for send

Solution

1. Step 1: Understand channel blocking behavior

   Sending on an unbuffered channel blocks until a receiver is ready.

2. Step 2: Analyze code flow

   Send happens before receive, but no goroutine to receive yet, causing deadlock.

3. Step 3: Fix by sending in a goroutine

   Wrap send in go func() { ch <- 10 }() to allow receive to proceed.

4. Final Answer:

   Deadlock because send blocks without receiver; fix by using a goroutine for send -> Option D

5. Quick Check:

   Send blocks without receiver = deadlock [OK]

Hint: Send blocks if no receiver; use goroutine to avoid deadlock [OK]

Common Mistakes:

• Ignoring deadlock and expecting output
• Adding imports unnecessarily
• Thinking channel is nil

5. You want to receive from two channels ch1 and ch2 but only proceed when both have sent a value. Which approach correctly uses blocking behavior to achieve this?

hard

A. Use a buffered channel to store values from both channels

B. Use a select statement with both channels in one case

C. Receive from ch1 and ch2 sequentially: val1 := <-ch1; val2 := <-ch2

D. Receive from only one channel and ignore the other

Solution

1. Step 1: Understand blocking on multiple channels

   Receiving sequentially from both channels blocks until each sends a value.

2. Step 2: Evaluate options

   Select cannot wait for both simultaneously in one case. Ignoring one channel misses data. Buffered channel does not guarantee both received.

3. Final Answer:

   Receive from ch1 and ch2 sequentially: val1 := <-ch1; val2 := <-ch2 -> Option C

4. Quick Check:

   Sequential receive blocks until both values arrive [OK]

Hint: Receive from each channel one by one to wait for both [OK]

Common Mistakes:

• Using select to wait for both at once
• Ignoring one channel's value
• Assuming buffered channel solves blocking