小能豆

Hi Can someone pls help me explain this channel code in Golang

go

import "fmt"

func main() {
    cnp := make(chan func(), 10)
    for i := 0; i < 4; i++ {
        go func() {
            for f := range cnp {
                f()
            }
        }()
    }
    cnp <- func() {
        fmt.Println("Here")
    }
    fmt.Println("Here1")
}

I have a few questions which I couldnt think answers of

  1. How does function in a channel work? Is this some sort of recursion?
  2. What is the significance of running for loop 4 times
  3. Why is “Here” not being printed?

Went through docs and youtube channels


阅读 55

收藏
2023-12-20

共1个答案

小能豆

Let’s break down the code and address your questions:

import "fmt"

func main() {
    cnp := make(chan func(), 10)

    // Launching 4 goroutines, each having a for loop to receive functions from the channel.
    for i := 0; i < 4; i++ {
        go func() {
            for f := range cnp {
                f() // Execute the received function
            }
        }()
    }

    // Sending a function into the channel.
    cnp <- func() {
        fmt.Println("Here")
    }

    fmt.Println("Here1")
}
  1. How does the function in a channel work? Is this some sort of recursion?

  2. No, it’s not recursion. The code uses a goroutine (go func() {...}) to create a concurrent execution context. A function is created (func() { fmt.Println("Here") }) and sent to the channel cnp. The goroutines created in the loop are reading functions from the channel and executing them. It’s a way of passing around and executing functions concurrently.

  3. What is the significance of running the for loop 4 times?

  4. The loop is launching 4 goroutines, each with a for loop that listens for functions from the channel cnp. This is done to demonstrate concurrent execution. Multiple goroutines are running concurrently, processing functions sent to the channel.

  5. Why is “Here” not being printed?

  6. The issue here is that the main goroutine (the one executing main()) is not waiting for the spawned goroutines to complete. It sends a function to the channel and immediately prints “Here1” without waiting for the spawned goroutines to process the function from the channel. As a result, the program may exit before the spawned goroutines get a chance to execute the function.

To ensure that all spawned goroutines get a chance to execute the function, you might want to wait for them to finish using synchronization mechanisms like a sync.WaitGroup or a channel-based signaling mechanism.

Here’s an example using a sync.WaitGroup:

import (
    "fmt"
    "sync"
)

func main() {
    cnp := make(chan func(), 10)
    var wg sync.WaitGroup

    for i := 0; i < 4; i++ {
        wg.Add(1)
        go func() {
            defer wg.Done()
            for f := range cnp {
                f()
            }
        }()
    }

    cnp <- func() {
        fmt.Println("Here")
    }

    // Close the channel to signal that no more functions will be sent.
    close(cnp)

    // Wait for all goroutines to finish.
    wg.Wait()

    fmt.Println("Here1")
}

This ensures that the main goroutine waits for the spawned goroutines to finish before printing “Here1”.

2023-12-20