问题

package main

import (
	"fmt"
)

func main() {
	arr0 := []int{
		1,2,3,4,5,
	}
	arr1 := []int{}

	fmt.Println(arr0)
	fmt.Println(arr1)
	fmt.Println("转移中...")
	transfer(&arr0, &arr1)
	fmt.Println(arr0)
	fmt.Println(arr1)
}

func transfer(arr0 *[]int, arr1 *[]int) {
	tmp := make([]int, 0)
	for i:=0;i<len(*arr0);i++ {
		tmp = append(tmp, (*arr0)[i])
	}

	*arr1 = tmp
	s := make([]int, 0)
	*arr0 = s
}

对于transfer函数，我打算将切片arr0的元素转移到切片arr1中，并清空切片arr0

但是这个操作没有成功

这是我的输出结果

[1 2 3 4 5]
[]
转移中...
[1 2 3 4 5]
[]

转移后，我需要以下结果。
[]
[1 2 3 4 5]
但实际上，main函数中的arr0和arr1保持不变！

有人能告诉我为什么不行吗？

我以为在内存中应该是这样的

[![enter image description here][1]][1]

运行transfer函数后

[![enter image description here][2]][2]

<details>
<summary>英文:</summary>

package main

import (
"fmt"
)

func main() {
arr0 := []int{
1,2,3,4,5,
}
arr1 := []int{}

fmt.Println(arr0)
fmt.Println(arr1)
fmt.Println("transferring...")
transfer(&arr0, &arr1)
fmt.Println(arr0)
fmt.Println(arr1)

}

func transfer(arr0 *[]int, arr1 *[]int) {
tmp := make([]int, 0)
for i:=0;i<len(*arr0);i++ {
tmp = append(tmp, (*arr0)[i])
}

arr1 = &tmp
s := make([]int, 0)
arr0 = &s

}

For function of transfer, I intented to transfer elements of slice arr0 to slice arr1 and empty slice arr0

But it is not successful

Here is my output

[1 2 3 4 5]
[]
transferring...
[1 2 3 4 5]
[]

After transferring, I need the result below. 
[]
[1 2 3 4 5]
But actually, arr0, and arr1 in the main function remain as it was!

can someone tell me why this is not ok?

I thought in the memory, it should be like this

[![enter image description here][1]][1]

after running transfer function

[![enter image description here][2]][2]


  [1]: https://i.stack.imgur.com/WUBOZ.png
  [2]: https://i.stack.imgur.com/e2TZa.png

</details>


# 答案1
**得分**: 3

@jacobsa给出了一个很好的答案。

记住这一点。你可以以更好的性能实现相同的效果。Golang为此提供了一个很好的机会。

```go
package main

import (
	"fmt"
)

func main() {
	arr0 := []int{
		1, 2, 3, 4, 5,
	}
	arr1 := []int{}

	fmt.Println(arr0)
	fmt.Println(arr1)
	fmt.Println("transferring...")
	transfer(&arr0, &arr1)
	fmt.Println(arr0)
	fmt.Println(arr1)
}

func transfer(arr0 *[]int, arr1 *[]int) {
	*arr0, *arr1 = *arr1, *arr0
}

package main

import (
	"fmt"
)

func main() {
	arr0 := []int{
		1, 2, 3, 4, 5,
	}
	arr1 := []int{}

	fmt.Println(arr0)
	fmt.Println(arr1)
	fmt.Println("transferring...")
	transfer(&arr0, &arr1)
	fmt.Println(arr0)
	fmt.Println(arr1)
}

func transfer(arr0 *[]int, arr1 *[]int) {
	*arr0, *arr1 = *arr1, *arr0
}

答案2

得分: 2

这两行代码：

arr1 = &tmp
arr0 = &s

在函数内部改变了局部变量arr1和arr0。这些变量恰好是指针，但它们只是main函数提供的输入指针的副本，它们不是指向输入指针的引用。

如果你改变了arr1和arr0指针所指向的内容，而不是指针本身，那么你会看到main函数提供的值发生变化：

*arr1 = tmp
*arr0 = s

arr1 = &tmp
arr0 = &s

*arr1 = tmp
*arr0 = s