I understand that the drop function is simply taking ownership of the passed value so Rust knows that once "drop" finishes, the _x can be 'dropped'.
But I though that the T had to be bound to have "Drop" trait (i.e. T: Drop) so that Rust knows it's possible to insert the call to 'drop' like _x.drop()?
The names are a bit confusing here, and I might petition to change them.
The `drop` here is just a simple library function and isn't technically related to the `Drop` trait, which is implemented with magical compiler pixie dust and allows you to define a destructor via a `.drop` method. Anything that goes out of scope has this `.drop` method called automagically.
You're correct in that if you were calling `x.drop()` explicitly, you would need to have a `Drop` bound on `T`.
If dropping was really implemented that way, every type would need to implement the Drop trait. Because otherwise there would be no way to free any memory at all.
Instead, Rust knows how to free any object. Implementing Drop is optional, and you only need to do it if you want some custom behavior at destruction.
pub fn drop<T>(_x: T) { }
I understand that the drop function is simply taking ownership of the passed value so Rust knows that once "drop" finishes, the _x can be 'dropped'.
But I though that the T had to be bound to have "Drop" trait (i.e. T: Drop) so that Rust knows it's possible to insert the call to 'drop' like _x.drop()?