如果对结果没有任何更改，为什么还要在以下脚本中添加＆？ [关闭]

Question

package main
import (
  "fmt"
  "math"
    "reflect"
)
type Vertex struct {
  X, Y float64
}
func (v *Vertex) Scale(f float64) {
  v.X = v.X * f
  v.Y = v.Y * f
}
func (v *Vertex) Abs() float64 {
  return math.Sqrt(v.X*v.X + v.Y*v.Y)
}
func main() {
  v := &Vertex{3, 4} // Whether or not with "&", the values don't change below.
  fmt.Printf("Before scaling: %+v, Abs: %v
", v, v.Abs())
  v.Scale(5)
  fmt.Printf("After scaling: %+v, Abs: %v
", v, v.Abs())
    fmt.Println(reflect.TypeOf(Vertex{3,4}))
}

---

Hello, I am learning golang now. I do not understand what is the use of adding "&", if it does not make any change on the result value?

I thought we add "&" to variables to get the memory address. If we can add "&" to Vertex{3,4}, does this mean it is variable? Confused.

Answer 1

I assume you're talking about Vertex vs &Vertex? Yes, adding & means that v now contains an address to a struct of type Vertex, whereas without the &, v would hold the struct directly.

In your example, using the address, or the struct directly, makes no difference. In many other cases, the distinction is very important.