What does the question mark mean in a type parameter bound?

Question

I found the definition for std::borrow::BorrowMut:

pub trait BorrowMut<Borrowed>: Borrow<Borrowed>
where
    Borrowed: ?Sized,
{
    fn borrow_mut(&mut self) -> &mut Borrowed;
}

What does the question mark in front of Sized mean in this type parameter bound (Borrowed: ?Sized)?

I consulted:

The Rust Programming Language¹ book,
The Rust Reference², and also
What does "Sized is not implemented" mean? on Stack Overflow

but didn't find an explanation. Please give a reference in your answer.

¹ _{especially see section 5.20 Traits}
² _{and section 6.1.9 Traits}

Shepmaster Shepmaster · Accepted Answer · 2015-05-19T18:44:58

It means that the trait is optional. The current syntax was introduced in the DST syntax RFC.

The only trait I am aware of that works for ? is Sized.

In this specific example, we can implement BorrowMut for unsized types, like [T] — note that there's no & here!

One built-in implementation makes use of that:

impl<T> BorrowMut<[T]> for Vec<T>

As Matthieu M. adds:

This is a case of a widening bound; in general bounds impose more constraints, but in the case of Sized it was decided that unless otherwise noted a generic T would be assumed to be Sized. The way to note the opposite would be to mark it ?Sized ("maybe Sized").

What does the question mark mean in a type parameter bound?

1 Answers