Buffered and unbuffered channels in Go - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of unbuffered channel communication

What is the output of this Go program using an unbuffered channel?

Go

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

A 42

B deadlock

C 0

D compile error

💡 Hint

Think about how unbuffered channels block until both sender and receiver are ready.

The unbuffered channel blocks the goroutine sending 42 until the main goroutine receives it. So the program prints 42.

❓ Predict Output

intermediate

Output of buffered channel with multiple sends

What will this Go program print?

Go

package main
import (
	"fmt"
)
func main() {
	ch := make(chan int, 2)
	ch <- 10
	ch <- 20
	fmt.Println(<-ch)
	fmt.Println(<-ch)
}

A 20\n10

B 10\n20

C deadlock

D compile error

💡 Hint

Buffered channels allow sending without immediate receiving until full.

The buffered channel holds two values. The first receive gets 10, the second gets 20.

❓ Predict Output

advanced

Deadlock detection with unbuffered channel

What happens when this Go program runs?

Go

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

A compile error

B prints 1

C compiles but no output

D deadlock runtime panic

💡 Hint

Consider that sending on an unbuffered channel blocks until a receiver is ready.

The send blocks forever because no goroutine is receiving yet, causing a deadlock panic.

❓ Predict Output

advanced

Channel capacity and length after sends

What will be the output of this Go program?

Go

package main
import (
	"fmt"
)
func main() {
	ch := make(chan int, 3)
	ch <- 5
	ch <- 10
	fmt.Println(len(ch), cap(ch))
}

A 2 3

B 3 3

C 0 3

D compile error

💡 Hint

len returns the number of elements in the channel buffer, cap returns its capacity.

Two values are sent, so len is 2. Capacity was set to 3.

🧠 Conceptual

expert

Behavior difference between buffered and unbuffered channels

Which statement correctly describes the difference between buffered and unbuffered channels in Go?

A Unbuffered channels store multiple values internally; buffered channels store only one value.

B Buffered channels always block the sender until the receiver receives; unbuffered channels never block.

C Unbuffered channels block the sender until the receiver receives; buffered channels allow sending without blocking until the buffer is full.

D Buffered channels do not support concurrent access; unbuffered channels do.

💡 Hint

Think about how sending and receiving synchronize in each channel type.

Unbuffered channels require sender and receiver to be ready simultaneously, causing blocking. Buffered channels allow sending up to their capacity without blocking.

Practice

1. What is the main difference between a buffered and an unbuffered channel in Go?

easy

A. Buffered channels allow sending multiple values without waiting, unbuffered channels require receiver ready.

B. Unbuffered channels store values internally, buffered channels do not.

C. Buffered channels block the sender until the receiver reads, unbuffered channels never block.

D. Unbuffered channels can only send strings, buffered channels can send any type.

Solution

1. Step 1: Understand unbuffered channel behavior

   Unbuffered channels require the sender to wait until the receiver is ready to receive the value.

2. Step 2: Understand buffered channel behavior

   Buffered channels have a capacity to hold multiple values, so the sender can send without waiting until the buffer is full.

3. Final Answer:

   Buffered channels allow sending multiple values without waiting, unbuffered channels require receiver ready. -> Option A

4. Quick Check:

   Buffered vs unbuffered = sender wait or not [OK]

Hint: Buffered = sender can send without waiting until full [OK]

Common Mistakes:

• Thinking unbuffered channels store values internally
• Believing buffered channels never block sender
• Confusing data types allowed in channels

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

easy

A. ch := make(chan int)

B. ch := make(chan int[3])

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

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

Solution

1. Step 1: Recall syntax for buffered channel creation

   Buffered channels are created with make(chan Type, capacity). Here, capacity is 3.

2. Step 2: Check each option

   ch := make(chan int, 3) uses make(chan int, 3) which is correct. ch := make(chan int) creates unbuffered channel. ch := make(chan int[3]) has invalid syntax. ch := new(chan int, 3) uses new which is incorrect for channels.

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 capacity creates unbuffered channel
• Using new() instead of make()
• Incorrect syntax with array notation

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

package main
import "fmt"
func main() {
  ch := make(chan int)
  go func() {
    ch <- 10
  }()
  fmt.Println(<-ch)
}

medium

A. Compilation error

B. 10

C. Deadlock error

D. 0

Solution

1. Step 1: Analyze channel type and goroutine

   The channel ch is unbuffered. The goroutine sends 10 to ch, and main receives from ch.

2. Step 2: Understand synchronization

   Because the send and receive happen in separate goroutines, the send blocks until receive is ready, so the program runs without deadlock and prints 10.

3. Final Answer:

   10 -> Option B

4. Quick Check:

   Unbuffered channel with goroutine send = prints sent value [OK]

Hint: Unbuffered channel blocks sender until receiver ready [OK]

Common Mistakes:

• Expecting deadlock without goroutine
• Thinking unbuffered channel stores default zero
• Confusing compilation error with runtime behavior

4. Consider this Go code snippet:

ch := make(chan int, 2)
ch <- 1
ch <- 2
ch <- 3
fmt.Println(<-ch)

What is the problem with this code?

medium

A. It causes a compilation error due to wrong syntax.

B. It prints 3 because the last value overwrites the buffer.

C. It compiles but prints 0 because channel is empty.

D. It causes a deadlock because the buffered channel is full before the third send.

Solution

1. Step 1: Check channel buffer capacity and sends

   The channel has capacity 2. Two sends fill the buffer. The third send blocks because buffer is full and no receiver yet.

2. Step 2: Understand blocking behavior

   Since no goroutine is receiving before the third send, the program deadlocks at third send.

3. Final Answer:

   It causes a deadlock because the buffered channel is full before the third send. -> Option D

4. Quick Check:

   Buffered channel full + send without receive = deadlock [OK]

Hint: Buffered channel blocks sender when full without receiver [OK]

Common Mistakes:

• Assuming buffer overwrites old values
• Expecting zero printed from empty channel
• Confusing syntax error with runtime deadlock

5. You want to send 5 integers to a channel without blocking the sender immediately. Which of these code snippets correctly achieves this using buffered channels?

Option A:
ch := make(chan int, 5)
for i := 1; i <= 5; i++ {
  ch <- i
}

Option B:
ch := make(chan int)
for i := 1; i <= 5; i++ {
  ch <- i
}

Option C:
ch := make(chan int, 3)
for i := 1; i <= 5; i++ {
  ch <- i
}

Option D:
ch := make(chan int, 0)
for i := 1; i <= 5; i++ {
  ch <- i
}

hard

A. Option A only

B. Option B only

C. Option C only

D. Option D only

Solution

1. Step 1: Understand buffered channel capacity

   To send 5 integers without blocking immediately, the channel buffer must be at least 5.

2. Step 2: Analyze each snippet's buffer size

   The first snippet creates a channel with capacity 5, allowing all 5 sends without blocking. The second creates an unbuffered channel, blocking the sender immediately on the first send. The third has capacity 3, blocking on the fourth send. The fourth has capacity 0 (unbuffered), blocking immediately.

3. Final Answer:

   Option A only -> Option A

4. Quick Check:

   Buffer size >= sends to avoid blocking [OK]

Hint: Buffer size must match sends to avoid blocking [OK]

Common Mistakes:

• Using unbuffered channels expecting no blocking
• Choosing smaller buffer than sends
• Misunderstanding buffer size zero means unbuffered