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：
type node struct {
identifier string
parent *node
children []*node
root int
}
func visitNodes(root *node) {
for i := 0; i < len(root.children); i++ {
fmt.Printf("Visiting node %s\n", &root.children[i])
printNodeAddress(root.children[i])
}
}
func printNodeAddress(node *node) {
fmt.Println(&node)
}
func main() {
root := new(node)
node1 := new(node)
node2 := new(node)
root.children = append(root.children, node1)
root.children = append(root.children, node2)
visitNodes(root)
}
产生的结果是：
Visiting node %!s(**main.node=0x10500170)
0x10500180
Visiting node %!s(**main.node=0x10500174)
0x10500190
而我期望它产生类似这样的结果：
Visiting node %!s(**main.node=0x10500170)
0x10500170
Visiting node %!s(**main.node=0x10500174)
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:
type node struct {
identifier string
parent *node
children []*node
root int
}
func visitNodes(root *node) {
for i := 0; i < len(root.children); i++ {
fmt.Printf("Visiting node %s\n", &root.children[i])
printNodeAddress(root.children[i])
}
}
func printNodeAddress(node *node) {
fmt.Println(&node)
}
func main() {
root := new(node)
node1 := new(node)
node2 := new(node)
root.children = append(root.children, node1)
root.children = append(root.children, node2)
visitNodes(root)
}
Produces:
Visiting node %!s(**main.node=0x10500170)
0x10500180
Visiting node %!s(**main.node=0x10500174)
0x10500190
While I expect it to produce something like this:
Visiting node %!s(**main.node=0x10500170)
0x10500170
Visiting node %!s(**main.node=0x10500174)
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：
func printNodeAddress(node *node) {
fmt.Println(&node) // 这里有一个额外的间接层级，一个 **node
}
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：
fmt.Printf("Visiting node %p\n", root.children[i])
并将你的 printNodeAddress 函数改为：
fmt.Printf("%p\n", node)
然后你会得到以下结果：
Visiting node 0x1052f2c0
0x1052f2c0
Visiting node 0x1052f2e0
0x1052f2e0
英文:
The issue is that you're taking the address of a pointer:
func printNodeAddress(node *node) {
fmt.Println(&node) // there's now a second layer of indirection in here. a **node
}
When really what you're trying to see is the pointer's memory address. You should change your Printf to this:
fmt.Printf("Visiting node %p\n", root.children[i])
And your printNodeAddress function to this:
fmt.Printf("%p\n", node)
Then you'll get this:
Visiting node 0x1052f2c0
0x1052f2c0
Visiting node 0x1052f2e0
0x1052f2e0
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