英文:
how to obtain a reference to an interface value
问题
可以在不使用反射的情况下获取接口值的引用吗？如果不能，为什么不能？
我尝试了以下代码：
package foo
type Foo struct {
a, b int
}
func f(x interface{}) {
var foo *Foo = &x.(Foo)
foo.a = 2
}
func g(foo Foo) {
f(foo)
}
但是它会报错：
./test.go:8: cannot take the address of x.(Foo)
英文:
Is it possibile to obtain a reference to an Interface Value without
reccurring to reflection? If not, why not?
I have attempted:
package foo
type Foo struct {
a, b int
}
func f(x interface{}) {
var foo *Foo = &x.(Foo)
foo.a = 2
}
func g(foo Foo) {
f(foo)
}
but it fails with:
./test.go:8: cannot take the address of x.(Foo)
答案1
得分: 1
清除你的疑惑，如果你按照"Assert"的意思来理解的话：
在你的例子中，x.(Foo)只是一种类型断言，它不是一个对象，所以你不能获取它的地址。
所以在运行时，当对象被创建时，例如：
var c interface{} = 5
d := c.(int)
fmt.Println("Hello", c, d) // Hello 5 5
它只是断言：
断言c不是nil，并且c中存储的值是int类型
所以它不是内存中的任何物理实体，但在运行时，d将根据断言的类型分配内存，然后将c的内容复制到该位置。
所以你可以这样做：
var c interface{} = 5
d := &c
fmt.Println("Hello", (*d).(int)) // hello 5
希望我解决了你的困惑。
英文:
To clear your doubts if you go by the meaning Assert
> state a fact or belief confidently and forcefully
in your example x.(Foo) is just a type assertion it's not a object so you can not get its address.
so at runtime when the object will be created for example
var c interface{} = 5
d := c.(int)
fmt.Println("Hello", c, d) // Hello 5 5
it only assert that
> asserts that c is not nil and that the value stored in c is of type int
So its not any physical entity in memory but at runtime d will be allocated memory based on asserted type and than the contents of c will be copied into that location.
So you can do something like
var c interface{} = 5
d := &c
fmt.Println("Hello", (*d).(int)) // hello 5
Hope i cleared your confusion.
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