go - goroutine 没有输出

Question

而SayHello()按预期执行，goroutine 什么也不打印。

package main

import "fmt"

func SayHello() {
    for i := 0; i < 10 ; i++ {
        fmt.Print(i, " ")
    }
}

func main() {
    SayHello()
    go SayHello()
}

Answer 1

当您的 main()函数结束，你的程序也结束。它不会等待其他 goroutine 完成。

引自 Go Language Specification: Program Execution :

Program execution begins by initializing the main package and then invoking the function main. When that function invocation returns, the program exits. It does not wait for other (non-main) goroutines to complete.

见 this answer更多细节。

你必须告诉你的 main()函数等待 SayHello()函数开始作为一个 goroutine 来完成。您可以将它们与 channel 同步，例如:

func SayHello(done chan int) {
    for i := 0; i < 10; i++ {
        fmt.Print(i, " ")
    }
    if done != nil {
        done <- 0 // Signal that we're done
    }
}

func main() {
    SayHello(nil) // Passing nil: we don't want notification here
    done := make(chan int)
    go SayHello(done)
    <-done // Wait until done signal arrives
}

另一种选择是通过关闭 channel 来表示完成:

func SayHello(done chan struct{}) {
    for i := 0; i < 10; i++ {
        fmt.Print(i, " ")
    }
    if done != nil {
        close(done) // Signal that we're done
    }
}

func main() {
    SayHello(nil) // Passing nil: we don't want notification here
    done := make(chan struct{})
    go SayHello(done)
    <-done // A receive from a closed channel returns the zero value immediately
}

备注:

根据您的编辑/评论:如果您希望 2 运行 SayHello()随机打印“混合”数字的功能:您无法保证观察到这种行为。再次，请参阅 aforementioned answer更多细节。 Go Memory Model仅保证某些事件在其他事件之前发生，您无法保证如何执行 2 个并发 goroutine。

您可能会尝试它，但要知道结果不会是确定性的。首先，您必须启用多个事件的 goroutine 来执行:

runtime.GOMAXPROCS(2)

其次你必须先开始 SayHello()作为 goroutine，因为您当前的代码首先执行 SayHello()在主 goroutine 中，只有在它完成后才会启动另一个:

runtime.GOMAXPROCS(2)
done := make(chan struct{})
go SayHello(done) // FIRST START goroutine
SayHello(nil) // And then call SayHello() in the main goroutine
<-done // Wait for completion