如何在函数中传递一个指向结构体的指针?

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英文:

How do I pass a pointer to a structure in a function?

问题

我想知道如何用什么来替换*Type?结构体在内部的地址是什么?

//mycode.go

package main

import "fmt"

func out(k *Type) {
    fmt.Println(k)
}

func main() {

    type DataIP struct{ Title, Desc string }

    Data := DataIP{
        "Hello!",
        "Hello GO!",
    }
    out(&Data)
}

我想知道如何用什么来替换*Type?结构体在内部的地址是什么?

英文:

I wonder how to replace *Type by ? What address has the structure inside?

//mycode.go

package main

import "fmt"

func out(k *Type) {
    fmt.Println(k)
}

func main() {

    type DataIP struct{ Title, Desc string }

    Data := DataIP{
        "Hello!",
        "Hello GO!",
    }
    out(&Data)
}

答案1

得分: 1

我不确定是否理解你的问题。

如果你希望out函数只能用于DataIP类型的结构体

只需在main函数外定义DataIP,并使用func out(k *DataIP)的函数签名。


如果你希望能够将任何类型的结构体传递给out函数

在Go语言中,可以使用接口类型来实现这种类型的通用方法。正如这个答案所解释的那样,接口是一个包含两个数据字的容器:

  1. 一个字用于指向值的底层类型的方法表,
  2. 另一个字用于指向该值所持有的实际数据。

接口可以持有任何类型的值,并且通常用作函数参数,以便能够处理多种类型的输入。

在你的情况下,你可以这样做:

func out(k interface{}) {
    fmt.Println(k)
}

这将打印出&{Hello! Hello GO!}。如果你希望&消失(即你总是传递指针),你可以使用reflect包来"取消引用" k

func out(k interface{}) {
    fmt.Println(reflect.ValueOf(k).Elem())
}

这将输出{Hello! Hello GO!}

这里有一个示例供你参考。


如果你想打印Data的地址

你可以使用fmt.Printf%p格式:

fmt.Printf("%p", &Data) // 0x1040a130

使用out函数,你可以这样做:

func out(k interface{}) {
    fmt.Printf("%p\n", k)
}

参考这个示例

英文:

I am not sure to understand your question.

If you want out to work only with structs of type DataIP:

simply define DataIP outside of main and use the signature func out(k *DataIP).


if what you want is to be able to pass any type of structure to out:

In golang, this sort of generic methods can be implemented using the interface type. As this answer explains, an interface is a container with two words of data:

  1. one word is used to point to a method table for the value’s underlying type,
  2. and the other word is used to point to the actual data being held by that value.

An interface can hold anything and is often used as a function parameter to be able to process many sort of inputs.

In your case, you can do:

func out(k interface{}) {
    fmt.Println(k)
}

This will print &{Hello! Hello GO!}. In case you want the & to disappear (i.e. you always pass it pointers), you can use the reflect package to "dereference" k:

func out(k interface{}) {
    fmt.Println(reflect.ValueOf(k).Elem())
}

which yields {Hello! Hello GO!}

Here is a playground example.


if what you want is to print the address of Data:

you can use the %p pattern with fmt.Printf:

fmt.Printf("%p", &Data) // 0x1040a130

Using the out function, you get:

func out(k interface{}) {
    fmt.Printf("%p\n", k)
}

See this playground example

答案2

得分: 1

你需要在main()之外定义类型DataIP,以便该类型在包的范围内而不仅仅在main函数内部:

package main

import "fmt"

type DataIP struct{ Title, Desc string }

func out(k *DataIP) {
    fmt.Println(k)
}

func main() {

    Data := DataIP{
        "Hello!",
        "Hello GO!",
    }
    out(&Data)
}

链接:https://play.golang.org/p/cUS6ttcUy-

英文:

You need to define the type DataIP outside of main() that the type is in the scope of the package and not just inside of the main function:

package main

import "fmt"

type DataIP struct{ Title, Desc string }

func out(k *DataIP) {
	fmt.Println(k)
}

func main() {

	Data := DataIP{
		"Hello!",
		"Hello GO!",
	}
	out(&Data)
}

https://play.golang.org/p/cUS6ttcUy-

huangapple
  • 本文由 发表于 2017年3月31日 17:04:46
  • 转载请务必保留本文链接:https://go.coder-hub.com/43136686.html
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