select 中尚未准备好向通道发送值

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

研究了一天还是无法解释为什么case ch <- 1:没有准备好，默认的case选择为运行。当然会造成死锁！

Answer 1

您已使用 make(chan int) 创建了未缓冲的 Go 通道。您需要一个缓冲通道（不一定会阻塞），您应该使用 make(chan int, 20) ，其中 20 是通道的大小

关于无缓冲通道的事情是它们也是同步的，所以它们总是在写入和读取时阻塞。当缓冲区填满时，缓冲通道也会发生同样的事情。

试试下面的代码

package main

import (
    "fmt"
    "time"
)

func main() {

    ch := make(chan int, 20)
    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)
}

Answer 2

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。这是首先生成的接收器 goroutine。此外，如果接收器尚未准备好，将选择默认情况。如果准备就绪，具体案例也将准备就绪。如果你运行多次这样，你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通道：

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)
}

此外，请务必在 Whosebug

上阅读此 thread

select 中尚未准备好向通道发送值

Sending value to a channel is not ready in the select

deadlock

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go

goroutine