理解Go指针

huangapple
go评论107阅读模式
英文:

Understanding Go Pointers

问题

我预期这段代码会生成4个地址，每个节点有2个地址，因此有2个相同的地址，然后另外一组2个相同的地址：

  1. type node struct {
  2. 	identifier string
  3. 	parent     *node
  4. 	children   []*node
  5. 	root       int
  6. }
  8. func visitNodes(root *node) {	
  9. 	for i := 0; i < len(root.children); i++ {
  10. 		fmt.Printf("Visiting node %s\n", &root.children[i])
  11. 		printNodeAddress(root.children[i])
  12. 	}
  13. }
  15. func printNodeAddress(node *node) {
  16. 	fmt.Println(&node)
  17. }
  20. func main() {
  21. 	root := new(node)
  22. 	node1 := new(node)
  23. 	node2 := new(node)
  25. 	root.children = append(root.children, node1)
  26. 	root.children = append(root.children, node2)
  27. 	
  28. 	visitNodes(root)
  29. }

产生的结果是：

  1. Visiting node  %!s(**main.node=0x10500170)
  2. 0x10500180
  3. Visiting node  %!s(**main.node=0x10500174)
  4. 0x10500190

而我期望它产生类似这样的结果：

  1. Visiting node  %!s(**main.node=0x10500170)
  2. 0x10500170
  3. Visiting node  %!s(**main.node=0x10500174)
  4. 0x10500174

我是否对Go指针的基本原理有误解，或者在处理切片时是否有所不同？

英文:

I expect this code to produce 4 addresses, 2 for each node, thus, having 2 identical addresses, and then another set of 2 identical addresses:

  1. type node struct {
  2. 	identifier string
  3. 	parent     *node
  4. 	children   []*node
  5. 	root       int
  6. }
  8. func visitNodes(root *node) {	
  9. 	for i := 0; i &lt; len(root.children); i++ {
  10.     	fmt.Printf(&quot;Visiting node %s\n&quot;, &amp;root.children[i])
  11.     	printNodeAddress(root.children[i])
  12. 	}
  13. }
  15. func printNodeAddress(node *node) {
  16. 	fmt.Println(&amp;node)
  17. }
  20. func main() {
  21. 	root := new(node)
  22. 	node1 := new(node)
  23. 	node2 := new(node)
  25. 	root.children = append(root.children, node1)
  26. 	root.children = append(root.children, node2)
  27. 	
  28. 	visitNodes(root)
  29. }

Produces:

  1. Visiting node  %!s(**main.node=0x10500170)
  2. 0x10500180
  3. Visiting node  %!s(**main.node=0x10500174)
  4. 0x10500190

While I expect it to produce something like this:

  1. Visiting node  %!s(**main.node=0x10500170)
  2. 0x10500170
  3. Visiting node  %!s(**main.node=0x10500174)
  4. 0x10500174

Am I misunderstanding the fundamentals of go pointers, or is different when dealing with slices?

答案1

得分: 5

问题在于你正在获取指针的地址：

  1. func printNodeAddress(node *node) {
  2.     fmt.Println(&node) // 这里有一个额外的间接层级，一个 **node
  3. }

实际上，你想要看到的是指针的内存地址。你应该将你的 Printf 改为：

  1. fmt.Printf("Visiting node %p\n", root.children[i])

并将你的 printNodeAddress 函数改为：

  1. fmt.Printf("%p\n", node)

然后你会得到以下结果：

  1. Visiting node 0x1052f2c0
  2. 0x1052f2c0
  3. Visiting node 0x1052f2e0
  4. 0x1052f2e0
英文:

The issue is that you're taking the address of a pointer:

  1. func printNodeAddress(node *node) {
  2.     fmt.Println(&amp;node) // there&#39;s now a second layer of indirection in here. a **node
  3. }

When really what you're trying to see is the pointer's memory address. You should change your Printf to this:

  1. fmt.Printf(&quot;Visiting node %p\n&quot;, root.children[i])

And your printNodeAddress function to this:

  1. fmt.Printf(&quot;%p\n&quot;, node)

Then you'll get this:

  1. Visiting node 0x1052f2c0
  2. 0x1052f2c0
  3. Visiting node 0x1052f2e0
  4. 0x1052f2e0

huangapple
  • 本文由 发表于 2013年12月13日 01:33:53
  • 转载请务必保留本文链接：https://go.coder-hub.com/20550274.html
  • go
  • pointers
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