Can golang not use pointers?

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Release: 2019-12-17 14:26:06
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Can golang not use pointers?

#Sometimes when I look at other people's Go code, I find that some of them use pointers in the code, and some don't. (Recommended learning: go)

Suppose there is a structure type called Person, and it is found that some methods use func methodA (*person Person) as a parameter, or use func *( person Person) methodA() is used as the structure's own method, that is, the person structure can directly call methodA, but using a pointer.

Or you can see the usage of var personMap map[string]*Person in the map structure

If you transfer from java to golang, it may not be easy to understand. Because there are no pointers in the Java world, you can just pass it directly and use it. However, you need to pay attention to many things when going to golang.

So when should you use it? Why is it needed in some places?

If you don’t use pointers, you cannot assign values ​​to the structure in some cases. Let’s look at a piece of code. This code does not use any pointers and first defines a bunch of objects for testing.

type Person struct {  //person结构体,包含年龄,名称,车
    age  int
    name string
    car  Car
}

type Car struct {  //person名下的车
    name string  //车的名字
}

var personMap map[string]Person   //一个存放person的map

func setName(person Person, name string) { //给参数person设置名字
    person.name = name
}

func (person Person) setName(name string) {  //设置名字
    person.name = name
}
func printName(person Person){  //打印person的名字
    fmt.Println(person.name)
}
func (person Person)printName(){  //结构体person自己支持打印名字
    fmt.Println(person.name)
}
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Then write the main method. I will comment the printed results in the code. I can find that the assignment fails in many cases.

func main() {
    person := Person{}
    fmt.Println(person)  //{0  {}}
    person.age = 12
    person.name = "小明"
    person.car = Car{"宝马"}
    fmt.Println(person)  //{12 小明 {宝马}},正常赋值给person变量,因为这是在方法里面的变量
    setName(person, "小红")
    fmt.Println(person) //{12 小明 {宝马}},小红赋值失败,传递给setName方法的person没有赋值成功
    person.setName("小红")
    fmt.Println(person) //{12 小明 {宝马}},person自己setName,还是失败
    personMap = make(map[string]Person)
    personMap["test"] = person
    person = personMap["test"]
    person.name = "小红"
    fmt.Println(person) //{12 小红 {宝马}},从map中取出person,给小红赋值成功
    for _, value := range personMap { //遍历map
        fmt.Println(value)//{12 小明 {宝马}},打印的还是小明,而不是小红,说明上面personMap["test"]对象赋值失败
    }
}
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Next, change it to use a pointer

type Person struct {
    age  int
    name string
    car  Car
}
type Car struct {
    name string
}
var personMap map[string]*Person
func setName(person *Person, name string) {
    person.name = name
}
func (person *Person) setName(name string) {
    person.name = name
}
func printName(person Person){
    fmt.Println(person.name)
}
func (person Person)printName(){
    fmt.Println(person.name)
}
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Modify the main method and use the & address character

func main() {
    person := Person{}
    fmt.Println(person) //{0  {}}
    person.age = 12
    person.name = "小明"
    person.car = Car{"宝马"}
    fmt.Println(person) //打印{12 小明 {宝马}}
    setName(&person, "小红")
    fmt.Println(person) //{12 小红 {宝马}}, 成功赋值!
    person.setName("小黑") 
    fmt.Println(person) //{12 小黑 {宝马}}, 成功赋值!
    personMap = make(map[string]*Person)
    personMap["test"] = &person
    person = *personMap["test"]
    person.name = "小兰"
    fmt.Println(person) //{12 小兰 {宝马}},成功赋值!
    for _, value := range personMap {
        fmt.Println(value) //&{12 小兰 {宝马}},读取的也是正确的小兰
    }
}
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So we have To conclude, when we need to modify the variable content of the structure, the structure variable parameters passed in by the method need to use pointers , that is, when the address of the structure needs to modify the variables of the architecture in the map , you also need to use the structure address as the value of the map

If you just read the structure variable, you can pass the reference directly without using the pointer

*type The type variable here stores an address. This needs to be clear. You need to use &type to get the address.

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