关于 goroutine 调度问题

学习 goroutine 的时候看过这个例子，不太明白为什么会这样？请看代码： ``` // This sample program demonstrates how to create goroutines and // how the scheduler behaves. package main import ( "fmt" "runtime" "sync" ) // main is the entry point for all Go programs. func main() { // Allocate 1 logical processor for the scheduler to use. runtime.GOMAXPROCS(1) // wg is used to wait for the program to finish. // Add a count of two, one for each goroutine. var wg sync.WaitGroup wg.Add(2) fmt.Println("Start Goroutines") // Declare an anonymous function and create a goroutine. go func() { // Schedule the call to Done to tell main we are done. defer wg.Done() // Display the alphabet three times for count := 0; count < 3; count++ { for char := 'a'; char < 'a'+26; char++ { fmt.Printf("%c ", char) } } }() // Declare an anonymous function and create a goroutine. go func() { // Schedule the call to Done to tell main we are done. defer wg.Done() // Display the alphabet three times for count := 0; count < 3; count++ { for char := 'A'; char < 'A'+26; char++ { fmt.Printf("%c ", char) } } }() // Wait for the goroutines to finish. fmt.Println("Waiting To Finish") wg.Wait() fmt.Println("\nTerminating Program") } ``` 这段代码为什么始终先输出第二个 goroutine 的内容，即大写字母，再输出第一个 goroutine 的内容小写字母？ goroutine 不是无序的吗，为什么这个例子总是顺序一一，并且第二个 goroutine 先打印？