Questions
What is the difference between a buffered and an unbuffered channel in Go?
The Scenario
You are a backend engineer at a fintech company. You are writing a new service that needs to communicate between two goroutines. You are not sure whether to use a buffered or an unbuffered channel.
The Challenge
Explain the difference between a buffered and an unbuffered channel in Go. What are the pros and cons of each approach, and which one would you choose for this use case?
A junior engineer might think that they are interchangeable. They might not be aware of the performance implications of choosing one over the other.
A senior engineer would be able to provide a detailed explanation of the differences between a buffered and an unbuffered channel. They would also be able to explain the trade-offs between each approach and would have a clear recommendation for which one to use in this use case.
Step 1: Understand the Key Differences
| Feature | Unbuffered Channel | Buffered Channel |
|---|---|---|
| Capacity | Has a capacity of zero. | Has a capacity of one or more. |
| Blocking | A sender is blocked until a receiver is ready, and a receiver is blocked until a sender is ready. | A sender is blocked only if the buffer is full, and a receiver is blocked only if the buffer is empty. |
| Use Cases | When you need to synchronize two goroutines. | When you want to decouple the sender and the receiver. |
Step 2: Choose the Right Tool for the Job
For our use case, it depends on whether we need to synchronize the two goroutines.
- If we need to make sure that the two goroutines are synchronized, we should use an unbuffered channel.
- If we want to decouple the two goroutines and allow them to run independently, we should use a buffered channel.
Step 3: Code Examples
Here are some code examples that show the difference between the two approaches:
Unbuffered Channel:
package main
import "fmt"
func main() {
ch := make(chan int)
go func() {
ch <- 1
}()
fmt.Println(<-ch)
}In this example, the main goroutine will be blocked until the other goroutine sends a value to the channel.
Buffered Channel:
package main
import "fmt"
func main() {
ch := make(chan int, 1)
ch <- 1
fmt.Println(<-ch)
}In this example, the main goroutine will not be blocked, because the channel has a capacity of one.
Practice Question
You have a sender goroutine that produces data at a faster rate than the receiver goroutine can consume it. Which of the following would be the most appropriate?