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過於簡化的 Golang 頻道！

WBOY原創: 2024-07-28 14:07:131089瀏覽

Oversimplified Golang Channel!

長話短說

本文解釋了 Go 通道，它可以實現 goroutine 之間的安全通訊。它涵蓋如何透過通道建立、傳送和接收數據，區分無緩衝和緩衝類型。它強調了關閉管道以防止僵局和改善資源管理的重要性。最後，介紹了用於高效管理多個通道操作的 select 語句。

Go Channel 簡介
建立頻道
傳送資料
接收資料
Go 中的通道類型
- 無緩衝通道
- 緩衝通道
關閉頻道
- 為什麼要關閉頻道？
不關閉頻道的程式碼片段
- 預期輸出與錯誤
關閉頻道的程式碼片段
- 預期輸出
使用 select 語句
- 使用通道進行選擇的範例
選擇常見問題
使用 WaitGroup 確保來自兩個通道的訊息
- 使用 WaitGroup 的範例
結論

Go 頻道簡介

Go，或 Golang，是一種功能強大的程式語言，旨在簡單和高效。它的突出特點之一是通道的概念，它促進了 goroutine 之間的溝通。通道允許安全的資料交換和同步，使並發程式設計更容易、更易於管理。

在本文中，我們將探索 Go 中的通道，分解它們的創建、資料傳輸和接收。這將幫助您了解如何在應用程式中有效地利用管道。

建立頻道

要在 Go 中建立通道，請使用 make 函數。這是一個簡單的程式碼片段，示範如何建立通道：

package main

import "fmt"

func main() {
    // Create a channel of type int
    ch := make(chan int)
    fmt.Println("Channel created:", ch)
}

在此範例中，我們建立一個可以發送和接收整數的通道 ch。預設情況下，通道是無緩衝的，這意味著它將阻塞，直到發送者和接收者都準備好。

當您執行提供的 Go 程式碼時，輸出將如下所示：

Channel created: 0xc000102060

解釋

頻道建立：
- 行 ch := make(chan int) 建立一個 int 類型的新通道。此通道可用於傳送和接收整數值。
頻道位址:
- 輸出0xc000102060是頻道的記憶體位址。在 Go 中，當您列印頻道時，它會顯示其內部表示，其中包括其在記憶體中的位址。
- 該位址指示通道在記憶體中的儲存位置，但它不提供有關通道狀態或內容的任何資訊。

傳送數據

建立通道後，您可以使用

go func() {
    ch <- 42 // Sending the value 42 to the channel
}()

在此片段中，我們啟動一個新的 goroutine，將整數值 42 送到通道 ch 中。這種非同步操作允許主程式在發送值時繼續執行。

接收資料

要從通道接收數據，您也可以使用

value := <-ch // Receiving data from the channel
fmt.Println("Received value:", value)

在此範例中，我們從通道 ch 讀取並將接收到的值儲存在變數 value 中。程式將阻塞在這一行，直到有值可供讀取。

Go 中的頻道類型

在 Go 中，通道主要分為兩種：無緩衝通道和緩衝通道。了解這些類型對於有效的並發程式設計至關重要。

1. 無緩衝通道

無緩衝通道是最簡單的類型。它沒有任何保存資料的能力；它要求發送者和接收者同時準備好。

特徵：

阻塞行為：發送和接收操作阻塞，直到雙方都準備好為止。這確保了 goroutine 之間的同步。
用例：最適合需要嚴格同步或通訊不頻繁的場景。

例子：

ch := make(chan int) // Unbuffered channel
go func() {
    ch <- 1 // Sends data; blocks until received
}()
value := <-ch // Receives data; blocks until sent
fmt.Println("Received:", value)

2. 緩衝通道

緩衝通道允許您指定容量，這表示它們在阻止發送之前可以保存有限數量的值。

Characteristics:

Non-blocking Sends: A send operation only blocks when the buffer is full. This allows for greater flexibility and can improve performance in certain scenarios.
Use Case: Useful when you want to decouple the sender and receiver, allowing the sender to continue executing until the buffer is filled.

Example:

ch := make(chan int, 2) // Buffered channel with capacity of 2
ch <- 1 // Does not block
ch <- 2 // Does not block
// ch <- 3 // Would block since the buffer is full
fmt.Println("Values sent to buffered channel.")

What is Closing a Channel?

In Go, closing a channel is an operation that signals that no more values will be sent on that channel. This is done using the close(channel) function. Once a channel is closed, it cannot be reopened or sent to again.

Why Do We Need to Close Channels?

Signal Completion: Closing a channel indicates to the receiving goroutine that no more values will be sent. This allows the receiver to know when to stop waiting for new messages.
Preventing Deadlocks: If a goroutine is reading from a channel that is never closed, it can lead to deadlocks where the program hangs indefinitely, waiting for more data that will never arrive.
Resource Management: Closing channels helps in managing resources effectively, as it allows the garbage collector to reclaim memory associated with the channel once it is no longer in use.
Iteration Control: When using a for range loop to read from a channel, closing the channel provides a clean way to exit the loop once all messages have been processed.

In this section, we will explore a Go code snippet that demonstrates the use of unbuffered channels. We will analyze the behavior of the code with and without closing the channel, as well as the implications of each approach.

Code Snippet Without Closing the Channel

Here’s the original code snippet without the close statement:

package main

import (
    "fmt"
)

func main() {
    messages := make(chan string)

    go func() {
        messages <- "Message 1"
        messages <- "Message 2"
        messages <- "Message 3"
        // close(messages) // This line is removed
    }()

    for msg := range messages {
        fmt.Println(msg)
    }
}

Expected Output and Error

fatal error: all goroutines are asleep - deadlock!

When you run this code, it will compile and execute, but it will hang indefinitely without producing the expected output. The reason is that the for msg := range messages loop continues to wait for more messages, and since the channel is never closed, the loop has no way of knowing when to terminate. This results in a deadlock situation, causing the program to hang.

Code Snippet With Closing the Channel

Now, let’s add the close statement back into the code:

package main

import (
    "fmt"
)

func main() {
    messages := make(chan string)

    go func() {
        messages <- "Message 1"
        messages <- "Message 2"
        messages <- "Message 3"
        close(messages) // Close the channel when done
    }()

    for msg := range messages {
        fmt.Println(msg)
    }
}

Expected Output

With the close statement included, the output of this code will be:

Message 1
Message 2
Message 3

Explanation of Closure Behavior

In this version of the code:

The close(messages) statement signals that no more messages will be sent on the messages channel.
The for msg := range messages loop can now terminate gracefully once all messages have been received.
Closing the channel allows the range loop to exit after processing all messages, preventing any deadlock situation.

Again, what if you don't close the channel?

Let's imagine a scenario where channels in Go are like people in a conversation.

Scene: A Coffee Shop

Characters:

Alice: Always eager to share ideas.
Bob: Takes a long time to respond.

Conversation:

Alice: "Hey Bob, did you hear about the new project? We need to brainstorm!"

Bob sips his coffee, staring blankly. The conversation is paused.

Alice: "Hello? Are you there?"

Bob looks up, still processing.

Bob: "Oh, sorry! I was... uh... thinking."

Minutes pass. Alice starts to wonder if Bob is even still in the chat.

Alice: "Should I keep talking or just wait for a signal?"

Bob finally responds, but it’s completely off-topic.

Bob: "Did you know that sloths can hold their breath longer than dolphins?"

Alice facepalms.

Alice: "Great, but what about the project?"

Bob shrugs, lost in thought again. The coffee shop becomes awkwardly silent.

Alice: "Is this conversation ever going to close, or will I just be here forever?"

Bob, now fascinated by the barista, mutters something about coffee beans.

Alice: "This is like a Go channel that never gets closed! I feel like I’m stuck in an infinite loop!"

Bob finally looks back, grinning.

Bob: "So... about those sloths?"

Moral of the Story: Sometimes, when channels (or conversations) don’t close, you end up with endless topics and no resolution—just like a chat that drags on forever without a conclusion!

Go Channels and the select Statement

Go's concurrency model is built around goroutines and channels, which facilitate communication between concurrent processes. The select statement is vital for managing multiple channel operations effectively.

Using select with Channels

Here's an example of using select with channels:

package main

import (
    "fmt"
    "time"
)

func main() {
    ch1 := make(chan string)
    ch2 := make(chan string)

    go func() {
        time.Sleep(1 * time.Second)
        ch1 <- "Result from channel 1"
    }()

    go func() {
        time.Sleep(2 * time.Second)
        ch2 <- "Result from channel 2"
    }()

    select {
    case msg1 := <-ch1:
        fmt.Println(msg1)
    case msg2 := <-ch2:
        fmt.Println(msg2)
    }
}

Output with select:

Result from channel 1

Why Does It Print Only One Output?

In Go, the select statement is a powerful construct used for handling multiple channel operations. When working with channels, you might wonder why a program prints only one output when multiple channels are involved. Let’s explore this concept through a simple example.

Scenario Overview

Consider the program that involves two channels: ch1 and ch2. Each channel receives a message after a delay, but only one message is printed at the end. You might ask, "Why does it only print one output?"

Timing and Concurrency

Channel Initialization: Both ch1 and ch2 are created to handle string messages.
Goroutines:
- A goroutine sends a message to ch1 after a 1-second delay.
- Another goroutine sends a message to ch2 after a 2-second delay.
Select Statement: The select statement listens for messages from both channels. It blocks until one of the channels is ready to send a message.

Execution Flow

When the program runs, it waits for either ch1 or ch2 to send a message.
After 1 second, ch1 is ready, allowing the select statement to execute the case for ch1.
Importantly, select can only execute one case at a time. Once a case is selected, it exits the select block.

FAQ on select

Q: Is it possible to wait for all channels in select to print all outputs?

A: No, the select statement is designed to handle one case at a time. To wait for multiple channels and print all outputs, you would need to use a loop or wait group.

Q: What happens if both channels are ready at the same time?

A: If both channels are ready simultaneously, Go will choose one at random to process, so the output may vary between executions.

Q: Can I handle timeouts with select?

A: Yes, you can include a timeout case in the select statement, allowing you to specify a duration to wait for a message.

Q: How can I ensure I receive messages from both channels?

A: To receive messages from both channels, consider using a loop with a select statement inside it, or use a sync.WaitGroup to wait for multiple goroutines to complete their tasks.

Ensuring Messages from Both Channels Using WaitGroup in Go

To ensure you receive messages from both channels in Go, you can use a sync.WaitGroup. This allows you to wait for multiple goroutines to complete before proceeding.

Here’s an example:

package main

import (
    "fmt"
    "sync"
    "time"
)

func main() {
    ch1 := make(chan string)
    ch2 := make(chan string)
    var wg sync.WaitGroup

    // Start goroutine for channel 1
    wg.Add(1)
    go func() {
        defer wg.Done()
        time.Sleep(1 * time.Second)
        ch1 <- "Result from channel 1"
    }()

    // Start goroutine for channel 2
    wg.Add(1)
    go func() {
        defer wg.Done()
        time.Sleep(2 * time.Second)
        ch2 <- "Result from channel 2"
    }()

    // Wait for both goroutines to finish
    go func() {
        wg.Wait()
        close(ch1)
        close(ch2)
    }()

    // Collect results from both channels
    results := []string{}
    for i := 0; i < 2; i++ {
        select {
        case msg1 := <-ch1:
            results = append(results, msg1)
        case msg2 := <-ch2:
            results = append(results, msg2)
        }
    }

    // Print all results
    for _, result := range results {
        fmt.Println(result)
    }
}

Output

Result from channel 1
Result from channel 2

Explanation

Channels and WaitGroup: Two channels, ch1 and ch2, are created. A sync.WaitGroup is used to wait for both goroutines to finish.
Goroutines: Each goroutine sends a message to its channel after a delay. The wg.Done() is called to signal completion.
Closing Channels: After all goroutines are done, the channels are closed to prevent any further sends.
Collecting Results: A loop with a select statement is used to receive messages from both channels until both messages are collected.
Final Output: The collected messages are printed.

This method ensures that you wait for both channels to send their messages before proceeding.

If you're interested in learning more about using sync.WaitGroup in Go, check out this article on concurrency: Golang Concurrency: A Fun and Fast Ride.

Real world example

Let's compare the two versions of a program in terms of their structure, execution, and timing.

Sequential Execution Version

This version processes the jobs sequentially, one after the other.

package main

import (
    "fmt"
    "time"
)

func worker(id int, job int) string {
    time.Sleep(time.Second) // Simulate work
    return fmt.Sprintf("Worker %d completed job %d", id, job)
}

func main() {
    start := time.Now()
    results := make([]string, 5)

    for j := 1; j <= 5; j++ {
        results[j-1] = worker(1, j) // Call the worker function directly
    }

    for _, result := range results {
        fmt.Println(result)
    }

    duration := time.Since(start)
    fmt.Printf("It took %s to execute!", duration)
}

Output:

Worker 1 completed job 1
Worker 1 completed job 2
Worker 1 completed job 3
Worker 1 completed job 4
Worker 1 completed job 5
It took 5.048703s to execute!

Concurrent Execution Version

This version processes the jobs concurrently using goroutines and channels.

package main

import (
    "fmt"
    "time"
)

func worker(id int, jobs <-chan int, results chan<- string) {
    for job := range jobs {
        time.Sleep(time.Second) // Simulate work
        results <- fmt.Sprintf("Worker %d completed job %d", id, job)
    }
}

func main() {
    start := time.Now()
    jobs := make(chan int, 5)
    results := make(chan string)

    for w := 1; w <= 3; w++ {
        go worker(w, jobs, results)
    }

    for j := 1; j <= 5; j++ {
        jobs <- j
    }
    close(jobs)

    for a := 1; a <= 5; a++ {
        fmt.Println(<-results)
    }

    duration := time.Since(start)
    fmt.Printf("It took %s to execute!", duration)
}