How to implement `Default` for a raw pointer?
rust, types
Solution
Is there a way to declare Default for the raw pointer?
No, currently there isn't. Either the trait or the type needs to be defined in the crate, in which the trait-impl is written (so called "orphan rules").
However, you don't need to manually implement `Default` for all of your types containing a pointer. You can create a new type, which wraps a raw pointer and does implement `Default`. Then you can just use this new type in all of your structs and simply derive `Default`.
struct ZeroedMutPtr<T>(pub *mut T);
impl<T> Default for ZeroedMutPtr<T> { ... }
Problem
When using raw points in a struct, Rust doesn't allow to derive from Default. eg: ``` #[derive(Default)] struct Foo { bar: *mut Foo, baz: usize, } ``` Reports ``` error[E0277]: the trait bound `*mut Foo: std::default::Default` is not satisfied ``` I tried this but it doesn't work: ``` impl Default for *mut Foo { fn default() -> *mut Foo { ptr::null_mut() } } ``` This gives an error: ``` impl doesn't use types inside crate ``` Is there a way to declare `Default` for the raw pointer? Otherwise I'll have to write explicit `default` functions for any `struct` which contains a raw pointer, OK in this example, but for larger structs it can be tedious, so I'd like to be able to avoid it in some cases.