5 ms·

Aren't unique pointer also scheduled for decommission?

• kibwen 13y ago
  No, not at all. Unique pointers are awesome and are the foundation for lots of other features. Now, some of the syntax for using them might still change. That's yet to be decided. But it wouldn't be Rust without unique pointers.
  • tarpden 13y ago
    Are "unique pointers" == "owned pointers"?
    • dbaupp 13y ago
      Yes, ~T.
    • Ygg2 13y ago
      Sorry about the confusion, I just translate ~T as a Uniq<T> because they are unique (i.e. you can't use them without cloning) and they were called like that on a mailing list a few times (IIRC).
• pcwalton 13y ago
  Undecided. They will always be somewhat magical in that the "box" operator creates them by default. But the syntax for them may change, or be moved into the library.
  • dded 13y ago
    You can put a pointer type into a library?
    • lucian1900 13y ago
      Rc and Gc are now libraries you can import. A nice advantage is that Arc, MutexArc and any other reference types look and work the same way.
    • dbaupp 13y ago
      Everything except the two reference types (& and &mut) can be trivially implemented in libraries: Rust is a low level language. E.g. the following is almost exactly identical to `~T` #[unsafe_no_drop_flag] struct Uniq<T> { priv ptr: *mut T } impl<T> Uniq<T> { fn new(value: T) -> Uniq<T> { unsafe { let ptr = malloc(size_of::<T>()) as *mut T; if ptr.is_null() { fail!("malloc failed"); } move_val_init(&mut *ptr, value); Uniq { ptr: ptr } } } fn borrow<'a>(&'a self) -> &'a T { unsafe {&*self.ptr} } fn borrow_mut<'a>(&'a mut self) -> &'a mut T { unsafe {&mut *self.ptr} } fn move(mut self) -> T { unsafe { let v = ptr::read_ptr(self.ptr as *T); free(self.ptr); self.ptr = 0; v } } } #[unsafe_destructor] impl<T> Drop for Uniq<T> { fn drop(&mut self) { unsafe { // this will free the pointer and run the // destructor of the inner type replace(self, transmute(0)).move(); } } }