https://qcrao91.gitbook.io/go/channel/channel-fa-song-he-jie-shou-yuan-su-de-ben-zhi-shi-shi-mo

Channel 发送和接收元素的本质是什么?

All transfer of value on the go channels happens with the copy of value.

就是说 channel 的发送和接收操作本质上都是 “值的拷贝”,无论是从 sender goroutine 的栈到 chan buf,还是从 chan buf 到 receiver goroutine,或者是直接从 sender goroutine 到 receiver goroutine。

举一个例子:

 
type user struct {
name string
age int8
}
 
var u = user{name: "Ankur", age: 25}
var g = &u
 
func modifyUser(pu *user) {
fmt.Println("modifyUser Received Vaule", pu)
pu.name = "Anand"
}
 
func printUser(u <-chan *user) {
time.Sleep(2 * time.Second)
fmt.Println("printUser goRoutine called", <-u)
}
 
func main() {
c := make(chan *user, 5)
c <- g
fmt.Println(g)
// modify g
g = &user{name: "Ankur Anand", age: 100}
go printUser(c)
go modifyUser(g)
time.Sleep(5 * time.Second)
fmt.Println(g)
}

运行结果:

 
&{Ankur 25}
modifyUser Received Vaule &{Ankur Anand 100}
printUser goRoutine called &{Ankur 25}
&{Anand 100}

这里就是一个很好的 share memory by communicating 的例子。

https://qcrao91.gitbook.io/go/channel/channel-fa-song-he-jie-shou-yuan-su-de-ben-zhi-shi-shi-mo
output

一开始构造一个结构体 u,地址是 0x56420,图中地址上方就是它的内容。接着把 &u 赋值给指针 g,g 的地址是 0x565bb0,它的内容就是一个地址,指向 u。

main 程序里,先把 g 发送到 c,根据 copy value 的本质,进入到 chan buf 里的就是 0x56420,它是指针 g 的值(不是它指向的内容),所以打印从 channel 接收到的元素时,它就是 &{Ankur 25}。因此,这里并不是将指针 g “发送” 到了 channel 里,只是拷贝它的值而已。

再强调一次:

Remember all transfer of value on the go channels happens with the copy of value.

 

 

参考:

https://qcrao91.gitbook.io/go/channel/channel-fa-song-he-jie-shou-yuan-su-de-ben-zhi-shi-shi-mo

https://www.cnblogs.com/bonelee/p/6056373.html

https://qcrao91.gitbook.io/go/channel/channel-fa-song-he-jie-shou-yuan-su-de-ben-zhi-shi-shi-mo

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