此Go代码如何在不取消引用的情况下通过指针设置对象的值？

Question

I'm learning Go from a Java/Python background, and am confused by this code from the Go tutorial. In the following code, the line

p.X = 1e9

sets the value of v.X to 1e9 using pointer p. As p is merely a pointer to v, isn't dereferencing necessary to set v's value? Thus the correct statement would be:

*p.X = 1e9 

Naturally, this results in an error. Can someone explain why the Go example code works as it is?

Code in question:

package main

import (
    "fmt"
)

type Vertex struct {
    X int
    Y int
}

func main() {
    v := Vertex{1, 2}
    p := &v
    p.X = 1e9
    fmt.Println(v)
}

Answer 1

In go, the compiler automatically converts the expression to (*p).X. From the the language spec:

if the type of x is a named pointer type and (*x).f is a valid selector expression denoting a field (but not a method), x.f is shorthand for (*x).f.