在Golang中将结构指针转换为接口指针 [英] Cast a struct pointer to interface pointer in Golang

问题描述

我有一个函数

  func doStuff（inout * interface {}）{
 ... 
} 
  

此函数的目的是能够将任何类型的指针视为输入。
但是当我想使用一个结构体的指针调用它时，我有一个错误。

  type MyStruct struct {
 f1 int 
} 
  

当调用 doStuff

  ms：= MyStruct {1} 
 doStuff（& ms）
  

我有

 $ c> test.go：38：不能在doStuff的参数中使用& ms（类型* MyStruct）作为类型** interface {} 
  

如何使& ms 与 * interface {}兼容 ？

解决方案

没有这样的东西作为指向接口的指针（技术上，你可以使用一个，不需要它。）

如：

  ms：= MyStruct {1} 
 doStuff（& ms）
 fmt.Printf（Hello，playground：％v \\\
，ms）
  

输出：

 您好，playground：{1} 
  

---

< a href =http://stackoverflow.com/users/86989/newacct> newacct mentions 在注释中：

将指针传递到接口直接工作，因为如果 MyStruct 符合协议，则 * MyStruct 也符合协议（因为类型的方法set包含在其指针类型的方法集中）。

在这种情况下，接口是空接口，所以它接受所有类型，但仍然。

I have a function

func doStuff(inout *interface{}) {
   ...
}

the purpose of this function is to be able to treat a pointer of any type as input. But when I want to call it with a the pointer of a struct I have an error.

type MyStruct struct {
    f1 int
}

When calling doStuff

ms := MyStruct{1}
doStuff(&ms)

I have

test.go:38: cannot use &ms (type *MyStruct) as type **interface {} in argument to doStuff

How can I cast &ms to be compatible with *interface{}?

解决方案

There is no such thing as a "pointer to an interface" (technically, you can use one, but generally you don't need it).

As seen in "what is the meaning of interface{} in golang?", interface is a container with two words of data:

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

So remove the pointer, and doStuff will work just fine: the interface data will be &ms, your pointer:

func doStuff(inout interface{}) {
   ...
}

See this example:

ms := MyStruct{1}
doStuff(&ms)
fmt.Printf("Hello, playground: %v\n", ms)

Output:

Hello, playground: {1}

---

As newacct mentions in the comments:

Passing the pointer to the interface directly works because if MyStruct conforms to a protocol, then *MyStruct also conforms to the protocol (since a type's method set is included in its pointer type's method set).

In this case, the interface is the empty interface, so it accepts all types anyway, but still.

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