当类型为 interface{} 时初始化 nil 指针

Question

我想知道如何初始化一个指向结构的nil指针，当函数接收此参数为.initializeinterface{}

假设我将始终发送一个指针，并且函数的参数类型严格，我如何使此代码工作？interface{}

type Foo struct {

}

func main() {

    var foo *Foo

    fmt.Println("foo is nil: ", foo)

    initialize(foo)

    fmt.Println("foo should not be nil: ", foo) // foo should be Foo{}, but it is nil

}

func initialize(fooPointer interface{}) {

    reflect.ValueOf(&fooPointer).Elem().Set(reflect.ValueOf(&Foo{}))

    fmt.Println("fooPointer is not nil: ", fooPointer)

}

https://play.golang.org/p/Va0bqXJGhWZ

Answer 1

若要修改被调用函数中的变量，必须将变量的地址传递给被调用函数。使用以下代码：func main() {&nbsp; &nbsp; var foo *Foo&nbsp; &nbsp; fmt.Println("foo is nil: ", foo)&nbsp; &nbsp; initialize(&foo)&nbsp; // <-- Pass address of variable&nbsp; &nbsp; fmt.Println("foo should not be nil: ", foo) // foo should be Foo{}, but it is nil}func initialize(fooPP interface{}) {&nbsp; &nbsp; // No need for & here because fooPP is a **Foo.&nbsp; &nbsp; reflect.ValueOf(fooPP).Elem().Set(reflect.ValueOf(&Foo{}))&nbsp; &nbsp; fmt.Println("fooPointer is not nil: ", fooPointer)}