How to move the concrete value out of a Box?

Question

I have value of type T in value: Box<dyn Any> and want to extract it. The only way I found is this:

let pv = value.downcast_mut::<T>();
let v = std::mem::replace(pv, T::default());

Is there a way to get v without requiring T to implement Default?

The suggested solution does not compile. – trent Jan 20 '20 at 13:33 — trent, Jan 20 '20 at 13:33

trent · Accepted Answer · 2020-01-20T14:08:01.377

Box has its own downcast which returns a Result<Box<T>, Box<dyn Any>>. Once you have a Box<T> you can simply dereference it to get the T out. Here's one way to use it:

fn get<T: Any>(value: Box<dyn Any>) -> T {
    let pv = value.downcast().expect("The pointed-to value must be of type T");
    *pv
}

How to move the concrete value out of a Box?

1 Answers1