多个goroutine的调度

亲爱的编程学习爱好者，如果你点开了这篇文章，说明你对《多个goroutine的调度》很感兴趣。本篇文章就来给大家详细解析一下，主要介绍一下，希望所有认真读完的童鞋们，都有实质性的提高。

package main

import "fmt"


func square(c chan int) {
    fmt.println("[square] reading (4)")
    num := <-c
    fmt.println("[square] calc (5)")
    c <- num * num
    fmt.println("back from [square] (10)")
}

func cube(c chan int) {
    fmt.println("[cube] reading (3)")
    num := <-c
    fmt.println("[cube] calc (11)")
    c <- num * num * num
    fmt.println("back from [cube] (12)")
}

func main() {
    fmt.println("[main] main() started (1)")

    squarechan := make(chan int)
    cubechan := make(chan int)

    go square(squarechan)
    go cube(cubechan)

    testnum := 3
    fmt.println("[main] sent testnum to squarechan (2)")

    squarechan <- testnum

    fmt.println("[main] resuming (6)")
    fmt.println("[main] sent testnum to cubechan (7)")

    cubechan <- testnum // why doesn't block here?

    fmt.println("[main] resuming (8)")
    fmt.println("[main] reading from channels (9)")

    squareval, cubeval := <-squarechan, <-cubechan
    fmt.println("[main] waiting calculating (13)")

    fmt.println("[main] results: ", squareval, cubeval)
    fmt.println("[main] main() stopped")
}

输出：

[main] main() started (1)
[main] sent testNum to squareChan (2)
[cube] reading (3)
[square] reading (4)
[square] calc (5)
[main] resuming (6)
[main] sent testNum to cubeChan (7)
[main] resuming (8)
[main] reading from channels (9)
back from [square] (10)
[cube] calc (11)
back from [cube] (12)
[main] waiting calculating (13)
[main] results:  9 27
[main] main() stopped

在上面给出的代码中，我认为 main() 例程应该在 cubechan <- testnum 之后被阻止，然后应该调度 cube 例程，这意味着输出 [cube] calc (11) 应该优先到 [main] 恢复 (8)。但在执行演示后，我对输出感到非常困惑。

谁能告诉我我是否误解了什么？

解决方案

它确实会阻塞，但在这些：

squareval, cubeval := <-squarechan, <-cubechan
                      ^             ^

如果 squarechan 或 cubechan 在其他例程中没有接收到任何值，则该行将陷入死锁，如下所示修改 cube() 以查看效果：

func cube(c chan int) {
    fmt.Println("[cube] reading (3)")
    _ = <-c
    fmt.Println("[cube] calc (11)")
    // c <- num * num * num
    fmt.Println("back from [cube] (12)")
}

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