先来一个返回指针的测试,结果跟想象一样

type A map[int]string
type B struct {
	A
	c int
}

func main() {
	b := B{make(A), 10}
	NewB := func() *B {
		return &b
	}
	c := NewB()
	c.c = 100
	c.A[1] = "3"
	fmt.Println(b, c)
}
/* output
{map[1:3] 100} &{map[1:3] 100}
*/

再试试直接返回值，这下的输出和想象中不一样了

type A map[int]string
type B struct {
	A
	c int
}

func main() {
	a := B{make(A), 10}
	b := B{make(A), 10}
	NewB := func() B {
		return b
	}
	c := NewB()
	c.c = 100
	c.A[1] = "3"
	// c[1]="3" invalid operation: c[1] (type B does not support indexing)
	fmt.Println(a, b, c)
}
/* output
{map[] 10} {map[1:3] 10} {map[1:3] 100}
*/

为什么呢，再试试

type A map[int]string
type B struct {
	A
	c int
}

func main() {
	a := make(A)
	NewA := func() A {
		return a
	}
	b := NewA()
	b[1] = "10"
	fmt.Println(a, b)
}
/* output
map[1:10] map[1:10]
*/
这样就编译不过了
func main() {
	a := make(A)
	NewA := func() *A {
		return &a
	}
	b := NewA()
	b[1] = "10" //invalid operation: b[1] (type *A does not support indexing)
	fmt.Println(a, b)
}

go 语言中map,slice都是指针类型，以值的形式返回struct的时候只是把map对应的指针拷贝了