Dereferencing Box gives back value instead of reference

Question

I can't seem to figure out why:

let a = Box::new(5i32);
let _:() = *a;

tells me that the assigned type on the second line is i32 and not &i32 since Deref.deref() (which I assume is being called at *a), returns &T.

Also, if I were to call deref() myself:

let _:() = <Box<i32> as Deref>::deref(&a);

I get the expected &i32.

Answer 1

Dereferencing doesn't necessarily produce an (intermediate) value. Consider

let b = Box::new(1);
(*b).clone();

The method i32::clone() is called with a &self argument where the reference points to the value inside the box, not to a temporary value that could be produced by (*b).

The trait Deref is part of implementing dereferencing (just like DerefMut).

There is no corresponding trait to what * can additionally do on a box: Move the inner value out and discard the box; this is colloquially called DerefMove but remains a compiler-hardcoded box speciality at this point.

When the compiler sees (*a), it has to infer whether to use Deref, DerefMut or “DerefMove”; it is inferred from how the expression is used: if you call a &self method on the result, Deref is used, for example.

Edited: Inherently copyable types (trait Copy), use Deref followed by copy instead of “DerefMove”; this is then no longer resticted to Box, but works with all smart pointers.

Answer 2

*foo is not the same as calling foo.deref(). If it were, you'd never be able to actually dereference a value. ^_^

* is syntactic sugar for "call Deref::deref and then go into the address provided". This is just what people expect from that operator. Deref exists to make it extensible.

That last bit glosses over some details that bluss' answer covers much better.

See also:

• How do I get an owned value out of a `Box`?

Answer 3

Box has an as_ref() implementation which simply returns a reference to the inner value:

let a = Box::new(5i32);
let _: &i32 = &a;