0

Is is possible for rust to have shallow copies, because it would appear that a move replaces a shallow copy?

roberthayek
  • 43
  • 1
  • 6
  • 6
    Trying to define a general concept of "shallow" and "deep" copy and then expecting a language to "support" this is a mistake in the first place. – Sebastian Redl Nov 12 '18 at 07:32

1 Answers1

1

You have to differentiate based on the underlying type.

You can't do a shallow copy of struct Foo(String); because two instances of Foo would point at the very same String, which would violate the strict aliasing rule .

However, if you are having a reference counter, e.g. struct Foo(Rc<String>); it is possible, because Rc will prevent you from doing unsafe things, e.g. having a mutable reference, when there are others references on the String.

It is also possible for types that implement [Clone] to make a "shallow" copy, because Copy implies, that the type can be copied by using memcpy (e.g. a u32 or a &T).

What is the difference between Copy and Clone? is also very worth reading.

So, in general no: Exceptions are reference counted structs (Rc or Arc) , Cloneables, or references (&T), because they don't violate the strict aliasing rule.

hellow
  • 10,920
  • 6
  • 47
  • 69
  • 2
    strict aliasing is not only about two pointer or reference to the same data, there also need to have the same type (expections types in C are character type). Put it simply you can't have a u32 * and a u16 * that refer to the same memory because the compiler assume it's not possible. Strict mean you can only have u32 * that refer the this memory (or char * in C), Strict != No aliasing. I don't know if it's true in Rust, but I will not take the risk, this kind of bug is hard to find. – Stargateur Nov 12 '18 at 09:43
  • Thanks for that. I would like to add, that safe rust won't let you do that, so there is no worry about that. – hellow Nov 12 '18 at 09:50
  • 1
    Rust does not have a strict aliasing rule for pointers. Even pointers of different types are allowed to refer to the same memory. Of course you need unsafe code to dereference them. – Sven Marnach Nov 12 '18 at 09:56