安排多个goroutine

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")
}

output:

[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

In the code given above, I think the main() routine should be blocked after cubeChan <- testNum, then cube routine should be scheduled, which means the output [cube] calc (11) ought to be prior to [main] resuming (8). But after executing on Playground, I'm so confused by the output.

Could anyone tell me if I misunderstand anything?

Answer 1

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

it does block, but at these:

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

if either squareChan or cubeChan doesn't receive any values in other routines, that line will become deadlock, modify cube() like below to see the effect:

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)")
}