[Arnavion]

>Can the allocator be customized?

`Box<T>` is the type of an owning pointer that uses the default global allocator, and `Box<T, A>` is the type of an owning pointer that uses an allocator of type `A`. The latter is unstable, ie it can only be used in nightly Rust.

(Also the fact that the latter changes the type means a large part of existing third-party code as well as a bunch of code in libstd itself becomes unusable if you want to use a type-level custom allocator because they only work with `Box<T>`. But that's a different discussion...)

>An object Y pointed by pY must live longer than object X.

Yes, the py field in Rust would use a reference type instead of a pointer, and the reference would need to have a lifetime annotation, and the compiler would work to prevent the situation you describe:

    struct X<'a> { py: &'a Y }

    let y = Y { ... };
    let x = X { py: &y };
    drop(y); // error: y is borrowed by x so it cannot be moved.

But to be clear, the `'a` lifetime syntax is not what's making this work. What's making this work is that the compiler tracks the lifetimes of references in general. This works in the same way even though there are no lifetime annotations:

    let y: String = "a".to_owned();
    let x = &y;
    drop(y); // error: y is borrowed by x so it cannot be moved.
    do_something_with(x);

The explicit lifetime annotations are just for a) readability, and b) because sometimes you want to name them to be able to express relationships between them. Eg if two lifetimes 'a and 'b are in play and you want to express that 'a is at least as long as 'b, then you have to write a `'a: 'b` bound. In many cases they can be omitted and the compiler infers them automatically.

[thradams]

(question about rust.. this is not implemented in cake yet)

Let's say I have to objects on the heap. A and B. A have a "view" to B.

Then we put a prompt for the user. (or dynamic condition) "Which object do you want to delete first A or B?" Then user select B. How this can be checked at compile time?

[Arnavion]

The code path that drops B will not compile unless that code path drops A first. It doesn't matter if that code path is in response to user input or not. Again, as I said, the point is that the compiler tracks the lifetime of all references. In this case A contains a reference to B, so any code that drops B without dropping A will not compile.