I don't understand your comment. `drop` takes a *mutable* reference. But by default, references in Rust are immutable.

littlestymaar 4 days ago | parent | next [-]

shared references are immutable in Rust, mut references and shared references are both references.

Ygg2 4 days ago | parent | prev | next [-]

Kibwen is saying that there was an idea to have `fn drop(&mut x)` from Drop trait become `fn drop_own(&own x)`. Then `&own` would do mutation and drop the owner of the reference. A regular &mut reference shouldn't do that outside of `drop(&mut x)`.

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codedokode 4 days ago | parent | prev [-]

Drop takes an object itself ("owned reference"), as I remember. Mutable ref allows reading/writing but not destroying or passing the ownership.

Owner = can read/write/destroy

Mutable ref = can read/write

Immutable ref = can only read, guarantered not to change

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steveklabnik 4 days ago | parent [-]

https://doc.rust-lang.org/stable/std/ops/trait.Drop.html#tym...

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codedokode 4 days ago | parent [-]

That's the destructor function, that is written by you and called by Rust before actually destroying something. The function that you want to look at is [1].

If you read the docs at your link it even says:

> This method is called implicitly when the value goes out of scope, and cannot be called explicitly (this is compiler error E0040).

> However, the mem::drop function in the prelude can be used to call the argument’s Drop implementation.

By the way the implementation of the function drop is just an empty function [2]; that's enough as local variables are destroyed on function return.

Mutable reference is a "borrow" which means you take a value from an owner and promise to return it back, and you cannot destroy a thing that you must return.

[1] https://doc.rust-lang.org/std/mem/fn.drop.html

[2] https://doc.rust-lang.org/src/core/mem/mod.rs.html#957

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steveklabnik 4 days ago | parent [-]

The drop function being talked about here is the one I pointed to, not the one you pointed to. The Drop trait is built into the language (as a lang item), std::mem::drop is just a regular old function.

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codedokode 4 days ago | parent [-]

The drop that you mention doesn't free memory, as I understand, it is called before actually destroying object's memory.

	▲	steveklabnik 4 days ago \| parent [-]
		Not inherently, sure. But ultimately this is far afield of what I was trying to say, which is that the signature being discussed of being changed from &mut T to &owned T is the one from the Drop trait. That’s it.