如何使用interface {}将整数作为指针传递给函数

Question

I thought this is a simple thing to do, but I was wrong. I can't pass integer as pointer to function using interface{}.

Example:

var test int
someChange(&test)
fmt.Printf("after function: %d", test)

func someChange(i interface{}) error{
    newFunnyValue := 42
    i = newFunnyValue
    fmt.Printf("hello from someChange now value test is: %d" , i)
    return nil //change gone ok, so nil
}

And result:

 hello from someChange now value test is: 42 
 after function: 0

I read that interface{} is similar to void* so above code should work but it's not, why? I want to add that if I pass some object which is a struct, everything works good.

Do I have to wrap int in some struct?

Edit:

https://play.golang.org/p/rg1vabug0P

Answer 1

If you want to observe the change outside of the someChange() function (in the test variable), you must modify the pointed value (assign a new value to it). You're not doing that, you just assign a new value to the i parameter (which is a local variable inside someChange()).

You may obtain the *int pointer from the i interface variable using type assertion, and then you can assign a new value to the pointed value.

Example:

func someChange(i interface{}) error {
    newFunnyValue := 42
    if p, ok := i.(*int); ok {
        *p = newFunnyValue
        return nil //change gone ok, so nil
    }
    return errors.New("Not *int")
}

Testing it:

var test int
someChange(&test)
log.Printf("after function: %d", test)

Output (try it on the Go Playground):

2009/11/10 23:00:00 after function: 42

Note that wrapping the int value or the *int pointer in a struct is unnecessary and it wouldn't make a difference if you're not assigning a new value to the pointed object.