go - 使用 channel 时出现问题 : all goroutines are asleep - deadlock

Question

我试图了解 channel 是如何运作的。我有这个例子:

import (
    "fmt"
    "runtime"
    "time"
)

func greet(c chan string) {
    fmt.Println("1: " + <-c)
    fmt.Println("2: " + <-c)
}

func main() {
    fmt.Println("main() started")

    c := make(chan string, 1)

    go greet(c)
    c <- "John"

    c <- "Mike"
    c <- "Mary"
    fmt.Println("active goroutines", runtime.NumGoroutine())
    c <- "Edward"
    time.Sleep(time.Second)
    fmt.Println("main() stopped")
}

在执行上述代码时，我看到它给出了一个错误“ fatal error :所有 goroutines 都处于 sleep 状态 - 死锁!”。但据我了解，在将“Edward”发送到 channel 后，执行应该会被阻止。

c <- "Edward"

这个程序应该打印出 Mary 和 Edward 的值。我也不会在任何地方关闭这个 channel 。谁能告诉我为什么，我该如何解决这个问题？

Answer 1

请引用以下代码，以便您更好地理解它。我也添加了评论。虽然，@mkopriva 消除了您的疑虑，但我认为如果我发布评论代码会更好。

package main

import (
    "fmt"
    "time"
)

func greet(c chan string) {
    // Range over the channel until close() is called
    for r := range c {
        // Print the names
        fmt.Println(r)
    }
}

func main() {
    fmt.Println("main() started")
    c := make(chan string, 1) // Make a bufferred channel with capacity of 1
    go greet(c)               // Spawn the goroutine
    c <- "John"               // Send
    c <- "Mike"               // Send
    c <- "Mary"               // Send
    c <- "Edward"             // Send
    close(c)                  // Close the channel which signals the for loop
    // inside greet to stop receiving
    time.Sleep(2 * time.Second) // Let us give the main gorutine some time
    // before it exits so that it lets the greet() to recieve all the names.
    // But it's better to use sync.WaitGroup for this.
    fmt.Println("main() stopped")
}