golang学习之指针、内存分配

golang学习之指针、内存分配

http://studygolang.com/articles/2658

 2015-03-18 13:39   neujie

 阅读 8760 次   0 人喜欢   0 条评论   收藏

1 func pointer_test() {
2     //空指针，输出为nil
3     var p *int
4     fmt.Printf("p: %v\n", p)
5     //指向局部变量，变量值初始为0
6     var i int
7     p = &i
8     fmt.Printf("p: %v,%v\n", p, *p)
9     //通过指针修改变量数值
10     *p = 8
11     fmt.Printf("p: %v,%v\n", p, *p)
12     //数组的初始化及输出
13     m := [3]int{3, 4, 5}
14     fmt.Printf("m:%v--%v,%v,%v\n", m, m[0], m[1], m[2])
15     //指针数组的初始化及输出
16     //j, k, l := 3, 4, 5
17     //x := [3]*int{&j, &k, &l}
18     x := [3]*int{&m[0], &m[1], &m[2]}
19     fmt.Printf("x:%v,%v,%v\n", x[0], x[1], x[2])
20     fmt.Printf("*x:%v,%v,%v\n", *x[0], *x[1], *x[2])
21
22     var n [3]*int
23     n = x
24     fmt.Printf("n:%v,%v,%v\n", n[0], n[1], n[2])
25     //指向数组的指针，也即二级指针的使用
26     y := []*[3]*int{&x}
27     fmt.Printf("y:%v,%v\n", y, y[0])
28     fmt.Printf("*y[0]:%v\n", *y[0])
29     fmt.Printf("*y[][]:%v,%v,%v\n", *y[0][0], *y[0][1], *y[0][2])
30     /*
31         *X            ->    v
32         [3]X        ->    [v0][v1][v2]
33         []X            ->    [v0]...[vi]
34         [3]*X        ->    [p0][p1][p2]
35                           |   |   |
36                           j   k   l
37         []*[3]*X    ->    [n0]...[ni]
38                           | ... |
39                          [p0]->j
40                          [p1]->k
41                          [p2]->l
42     */
43 }
44
45 type Student struct {
46     name  string
47     id    int
48     score uint
49 }
50
51 func memery_test() {
52     //new分配出来的数据是指针形式
53     p := new(Student)
54     p.name = "China"
55     p.id = 63333
56     p.score = 99
57     fmt.Println(*p)
58     //var定义的变量是数值形式
59     var st Student
60     st.name = "Chinese"
61     st.id = 666333
62     st.score = 100
63     fmt.Println(st)
64     //make分配slice、map和channel的空间，并且返回的不是指针
65     var ptr *[]Student
66     fmt.Println(ptr)     //ptr == nil
67     ptr = new([]Student) //指向一个空的slice
68     fmt.Println(ptr)
69     *ptr = make([]Student, 3, 100)
70     fmt.Println(ptr)
71     stu := []Student{{"China", 3333, 66}, {"Chinese", 4444, 77}, {"Chince", 5555, 88}}
72     fmt.Println(stu)
73 }

output:

{China 63333 99}

{Chinese 666333 100}

<nil>

&[]

&[{ 0 0} { 0 0} { 0 0}]

[{China 3333 66} {Chinese 4444 77} {Chince 5555 88}]