go - 多个goroutine的调度

Question

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()程序应在 cubeChan <- testNum 之后被阻止, 然后 cube应该安排例程，这意味着输出 [cube] calc (11)应该在 [main] resuming (8) 之前.但是在 Playground 上执行之后，我对输出感到很困惑。

如果我误解了什么，谁能告诉我？

Answer 1

I think the main() routine should be blocked after cubeChan <- testNum

它确实会阻塞，但在这些:

squareVal, cubeVal := <-squareChan, <-cubeChan
                      ^             ^

如果 squareChan 或 cubeChan 在其他例程中没有收到任何值，该行将成为死锁，将 cube() 修改为下面看看效果:

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)")
}