I run the following Go code 10 times:
package main
import (
"fmt"
"time"
)
func main() {
go fmt.Println("Hello")
fmt.Println("World")
time.Sleep(1 * time.Millisecond)
}
The output is always "World" first:
World
Hello
Does it manifest the goroutine will execute unitl there is a block in main routine? What is the execution time point of goroutine?
分享 What is the execution time point of goroutine?
The compiler will insert yield points into your program at different locations where it seems adequate. For example, in some function calls and seemingly tight loops. Also, a goroutine will yield when blocking on a syscall. So your program will probably yield execution over to the second goroutine when the first goroutine reaches fmt.Println("World") and enters a write syscall. After that non-determinism ensues because we don't know how long this syscall will take.
This, however, is an implementation detail and you should not be concerned with it. When talking about parallelism, the only timing guarantees (i.e. "happens before") you have are those provided by concurrency primitives like those from the standard library's sync/atomic package.
TL;DR: Don't rely on yield points for your program to work correctly.
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