理解代码-并发模式：菊花链

Question

I was studying Go concurrency pattern.

One pattern I am not sure is： Daisy Chain https://talks.golang.org/2012/concurrency.slide#39

It's very hard for me to understand the control flow of the code.

Can someone explain to me ?

package main

import (
    "fmt"
)

func f(left, right chan int) {
    left <- 1 + <-right
}

func main() {
    const n = 10000

    leftmost := make(chan int)
    right := leftmost               //point B: what does these do ?
    left := leftmost

    for i := 0; i < n; i++ {
        right = make(chan int)
        go f(left, right)
        left = right                //point A
    }
    go func(c chan int) { c <- 1 }(right)  
    fmt.Println(<-leftmost)
}

Conclusion:

1. the flow of channel going from right to left. It is good practice to write func f(left chan<- int, right <-chan int) rather than original function signature as above.

2. 'chain reaction' does not start until c <- 1, when signal 1 is sent to right most channel, reaction goes all the way to left most end. Print out 10001.

The reason is go channel block 'read' until received channel receive signal.

1. @Rick-777 shows how to use array like structure for easy understanding. Since each go coroutine is just around 6k big. It's not a bad idea to make 10k channel.

2. I clean up some code around Point B, for channel initialization. Here is the source code: http://play.golang.org/p/1kFYPypr0l

Answer 1

VonC has already given a direct answer. Here are some further remarks.

A slightly tidied-up version is in the playground, the difference being that the channels passed as parameters have their direction specified explicitly, ie. <-chan and chan<-. It's good practice to do this because the compiler can catch more mistakes for you.

An alternative and equivalent program that has a daisy-chain of n goroutines can be written using an array of channels instead. This allocates the same total number of channels using fewer lines of code. See playground:

package main

import (
    "fmt"
)

func f(left chan<- int, right <-chan int) {
    left <- 1 + <-right
}

func main() {
    const n = 10000

    // first we construct an array of n+1 channels each being a 'chan int'
    var channels [n+1]chan int
    for i := range channels {
        channels[i] = make(chan int)
    }

    // now we wire n goroutines in a chain
    for i := 0; i < n; i++ {
        go f(channels[i], channels[i+1])
    }

    // insert a value into the right-hand end
    go func(c chan<- int) { c <- 1 }(channels[n])

    // pick up the value emerging from the left-hand end
    fmt.Println(<-channels[0])
}

I hope you can see now how the original program is equivalent to this program. There is one minor difference: the original program does not create any channel array, so uses just a little less memory.