幾天前在 Golang Taiwan 的 slack 裡看到了一些關於 goroutine 的討論

有人問了以下程式碼 (from http://blog.mergermarket.it/now-youre-thinking-with-channels/)

package main

import "fmt"

func main() {
	channelForInts := make(chan int)
	go printIntFromChannel(channelForInts)
	channelForInts <- 5
}

func printIntFromChannel(channel chan int) {
	number := <-channel
	fmt.Println("The number is:", number)
}

提到上述程式碼在 goplayground 上測試好幾次都會印出 The number is:5

但他自己在本機跑是怎樣執行都不會出現 5

看到這個問題第一個想到的是 sync.Waitgroup

之前剛開始學 goroutine 的概念時，就在 stackoverflow 上看到有人討論相關的議題

加個 sync.Waitgroup 就可以保證看到 goroutine 執行完後的結果。

而事實上會這樣子的原因 Go Programming Language Specification 有寫

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.

所以在 main 執行完後也不管 goroutine 了，反正我就是要關掉了～這樣

改成用 sync.Waitgroup 的話大概就是長下面這樣，有稍微變形一下，多另外一個 goroutine 來往 channel 丟值

package main

import (
	"fmt"
	"sync"
)

var wg sync.WaitGroup

func main() {
	channelForInts := make(chan int)

	wg.Add(2)
	go setNum(channelForInts)
	go printIntFromChannel(channelForInts)

	wg.Wait()
}

func setNum(channel chan int) {
	defer wg.Done()
	
	for i := 0; i < 100; i++ {
		channel <- i
	}
	close(channel)
}

func printIntFromChannel(channel chan int) {
	defer wg.Done()

	for {
		number, ok := <-channel

		if !ok {
			break
		}

		fmt.Println("The number is:", number)
	}
}

另外社群朋友也提到另一個觀點，那就是將 GOMAXPROCS 設成 1

The GOMAXPROCS variable limits the number of operating system threads that can execute user-level Go code simultaneously. There is no limit to the number of threads that can be blocked in system calls on behalf of Go code; those do not count against the GOMAXPROCS limit.

且 goroutine 內沒有 system call，就不會有 context switch

所以會等 goroutine 跑完後回 main 把整個 process 完成

也是挺有趣的，從沒想過這個地方。

最後就是看起來這兩個來源值得一讀

1. Golang 的 spec https://golang.org/ref/spec
2. Golang Memory Model https://golang.org/ref/mem

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