What happens when I call std::mem::drop with a reference instead of an owned value?

Question

fn main() {
    let k = \"fire\";

    drop(k);

    println!(\"{:?}\", k);
}

Playground

Why am I

Answer 1

What happens when I call std::mem::drop with a reference

The reference itself is dropped.

a reference instead of an owned value

A reference is a value.

Why am I still able to use k after dropping it?

Because immutable pointers implement Copy. You pass in a copy of the reference and it's dropped.

Does drop not deref a reference automatically?

No, it does not.

what does the implementation of Drop look like for &str?

There isn't one for any kind of reference, immutable or mutable, so it's effectively ¹:

impl Drop for &str {
    fn drop(&mut self) {}
}

See also:

• Moved variable still borrowing after calling `drop`?
• Why does a mutable reference to a dropped object still count as a mutable reference?

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1 — As Peter Hall points out, there is a difference between having an empty Drop implementation and having no user-provided Drop implementation, but for the purposes of this question they are the same.