去渠道：为什么有两个不同的输出？

Question

I'm trying to understand channels in Go. Here is a code example:

package main

import "fmt"

func main() {
    m := make(map[int]string)
    m[2] = "First Value"
    c := make(chan bool)
    go func() {
        m[2] = "Second Value"
        c <- true
    }()
    fmt.Printf("1-%s
", m[2])
    fmt.Printf("2-%s
", m[2])
    _ = <-c
    fmt.Printf("3-%s
", m[2])
    fmt.Printf("4-%s
", m[2])
}

Sometimes the output of the above code was (result 1):

1-First Value
2-First Value
3-Second Value
4-Second Value

but sometimes I got (result 2):

1-First Value
2-Second Value
3-Second Value
4-Second Value

After changing c := make(chan bool) to c := make(chan bool, 1), the same occurred: sometimes result 1, sometimes result 2.

Why?

Answer 1

Your results makes perfect sense. As go routine run independent of each other, you would never know when go routine will start executing. As soon as line

m[2] = "Second Value"

is executed it will be reflected on your main go routine.

Hence when above line is executed between first and second print of your program you see result as

1-First Value
2-Second Value
3-Second Value
4-Second Value

When it is not you see other one. Before third print you ensure that other go routine is finished.

Just to clear it even more if you modify your program a little bit like

package main

import "fmt"
import "time"

func main() {
    m := make(map[int]string)
    m[2] = "First Value"
    c := make(chan bool)
    go func() {
        m[2] = "Second Value"
        c <- true
    }()
    time.Sleep(time.Second)
    fmt.Printf("1-%s
", m[2])
    fmt.Printf("2-%s
", m[2])
    _ = <-c
    fmt.Printf("3-%s
", m[2])
    fmt.Printf("4-%s
", m[2])
}

<kbd>Playground</kbd>

You will most probably get the following output

1-Second Value
2-Second Value
3-Second Value
4-Second Value

Hope it helps.