我在 Golang 教程中學(xué)習(xí) Go select 語(yǔ)句時(shí)嘗試對(duì)代碼進(jìn)行一些更改： https: //tour.golang.org/concurrency/5。但是，我遇到了問(wèn)題：fatal error: all goroutines are asleep - deadlock!goroutine 1 [chan send]:main.main()        concurrency.go:26 +0xa3goroutine 33 [chan receive]:main.main.func1(0xc000088000)        concurrency.go:24 +0x42created by main.main        concurrency.go:23 +0x89exit status 2這是我嘗試并遇到問(wèn)題的代碼func fibonacci(c, quit chan int) {    x, y := 0, 1    for {        select {            case c <- x:  //sending value x into channel c                x, y = y, x+y            case <-quit:    //receive value from quit                fmt.Println("quit")        return        }    }}func main() {    //create two channels    c := make(chan int)    quit := make(chan int)    go func() { //spin off the second function in order to let consume from c , so fibonaci can continue to work        fmt.Println(<-c)    //read value from channel c    }()    //Try moving the statement that send value to channel quit in order to     //return function fibonacci    quit <- 0       fibonacci(c, quit)}起初，我認(rèn)為結(jié)果將與下面代碼的結(jié)果相同//function fibonacci is same with the first onefunc fibonacci(c, quit chan int) {    x, y := 0, 1    for {        select {            case c <- x:  //sending value x into channel c                x, y = y, x+y            case <-quit:    //receive value from quit                fmt.Println("quit")        return        }    }}func main() {    //create two channels    c := make(chan int)    quit := make(chan int)    go func() { //spin off the second function in order to let consume from c , so fibonaci can continue to work        fmt.Println(<-c)    //read value from channel c        quit <- 0 //CHANGE: move the statement inside the closure function     }()    fibonacci(c, quit)}輸出是0quit您能解釋一下執(zhí)行第一個(gè)示例時(shí)死鎖的根本原因是什么嗎？在go例程中發(fā)送值退出通道與在主線程中發(fā)送值退出通道有什么區(qū)別？感謝你們。