对于结构顶点，map [int] vertex和map [int] * vertex有什么区别？

Question

To define a map from int to struct vertex, should I define map[int]vertex or map[int]*vertex? Which one is preferred?

I extended Chickencha's code:

package main

type vertex struct {
    x, y int 
}

func main() {
    a := make(map[int]vertex)
    b := make(map[int]*vertex)

    v := &vertex{0, 0}
    a[0] = *v
    b[0] = v 

    v.x, v.y = 4, 4
    println(a[0].x, a[0].y, b[0].x, b[0].y)

    //a[0].x = 3 // cannot assign to (a[0]).x
    //a[0].y = 3 // cannot assign to (a[0]).y
    b[0].x = 3 
    b[0].y = 3 
    println(a[0].x, a[0].y, b[0].x, b[0].y)

    u1 := a[0]
    u1.x = 2 
    u1.y = 2 
    u2 := b[0]
    u2.x = 2 
    u2.y = 2 
    println(a[0].x, a[0].y, b[0].x, b[0].y)
}

The output:

0 0 4 4
0 0 3 3
0 0 2 2

From the output, my understanding is, if I want to change the struct member in place, I must use pointer to the struct. But I'm still not sure the underlying reasons. Especially, why I cannot assign to a[0].x?

Answer 1

The main difference is that map[int]vertex stores vertex values and map[int]*vertex stores vertex references (pointers). The output of the following program should help illustrate:

package main

type vertex struct {
    x, y int
}

func main() {
    a := make(map[int]vertex)
    b := make(map[int]*vertex)

    v := &vertex{0, 0}
    a[0] = *v
    b[0] = v

    v.x, v.y = 4, 4
    println(a[0].x, a[0].y, b[0].x, b[0].y)
}

Output:

0 0 4 4

The vertex stored in b is modified by the v.x, v.y = 4, 4 line, while the vertex stored in a is not.