Are Go Channels Passed by Reference Implicitly?

Implicit Reference Passing in Go Channels

Contrarily to the expected behavior of copying data, channels in Go behave differently. Consider the following code, mentioned in Go's concurrency documentation:

package main

import "fmt"

func sum(a []int, c chan int) {
    sum := 0
    for _, v := range a {
        sum += v
    }
    c <- sum // send sum to c
}

func main() {
    a := []int{7, 2, 8, -9, 4, 0}

    c := make(chan int)
    go sum(a[:len(a)/2], c)
    go sum(a[len(a)/2:], c)
    x, y := <-c, <-c // receive from c

    fmt.Println(x, y, x+y)
}

In this example, the channel c is modified within the sum function. However, the changes persist after the function completes. This suggests that the channel is being passed by reference. However, no pointer to c was explicitly created.

So, the question arises: are channels passed by reference implicitly in Go?

Answer

Technically, channels are not passed by reference. Instead, Go makes a copy of the channel's internal data structure. This allows for efficient data communication between goroutines. However, since the built-in make function allocates memory on the heap, the created channel type behaves similarly to a reference type.

This means that you can modify the channel's content within a function, and those changes will still be visible to the calling function. Thus, channels can be used as reference types in Go programming, providing a convenient way to pass data between concurrently running goroutines.

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