go - 在选择中未准备好向 channel 发送值

Question

package main

import (
    "fmt"
    "time"
)

func main() {

    ch := make(chan int)
    go func() {
        fmt.Printf("func at %d\n", time.Now().UnixNano())
        select {
        case ch <- 1:
            fmt.Println("running send")
        default:
            fmt.Println("running default")
        }
    }()

    time.Sleep(100 * time.Millisecond)
    fmt.Printf("main at %d\n", time.Now().UnixNano())
    fmt.Println(<-ch)
}

the playground here

我研究了一天但仍然无法解释为什么 case ch <- 1:没有准备好，默认情况下选择运行。当然，它会导致死锁!

Answer 1

来自 doc :

If the channel is unbuffered, the sender blocks until the receiver has received the value. If the channel has a buffer, the sender blocks only until the value has been copied to the buffer; if the buffer is full, this means waiting until some receiver has retrieved a value.

一种方法是使用它。这是接收器 goroutine 首先生成的。此外，如果接收器尚未准备好，则会选择默认情况。如果它准备好了，具体案例也将准备好。如果您多次运行此程序，您可以看到任何一种情况发生。

package main

import (
    "fmt"
    "time"
)

func main() {
    ch := make(chan int)
    // goroutine acts as the reciever
    go func() {
        fmt.Printf("main at %d\n", time.Now().UnixNano())
        fmt.Println(<-ch)
    }()
    go func() {
        fmt.Printf("func at %d\n", time.Now().UnixNano())
        select {
        case ch <- 1:
            fmt.Println("running send")
        default:
            fmt.Println("running default")
        }
    }()
    time.Sleep(1 * time.Second) // Wait for the goroutines
}

另一种解决方案是使用缓冲 channel :

package main

import (
    "fmt"
    "time"
)

func main() {

    ch := make(chan int, 1)
    go func() {
        fmt.Printf("func at %d\n", time.Now().UnixNano())
        select {
        case ch <- 1:
            fmt.Println("running send")
        default:
            fmt.Println("running default")
        }
    }()

    time.Sleep(100 * time.Millisecond)
    fmt.Printf("main at %d\n", time.Now().UnixNano())
    fmt.Println(<-ch)
}

另外，请阅读本文 thread在堆栈溢出