rust - Does println! borrow or own the variable?

Question

I am confused with borrowing and ownership. In the Rust documentation about reference and borrowing

let mut x = 5;
{
    let y = &mut x;
    *y += 1;
}
println!("{}", x);

They say

println! can borrow x.

I am confused by this. If println! borrows x, why does it pass x not &x?

I try to run this code below

fn main() {
    let mut x = 5;
    {
        let y = &mut x;
        *y += 1;
    }
    println!("{}", &x);
}

This code is identical with the code above except I pass &x to println!. It prints '6' to the console which is correct and is the same result as the first code.

Answer 1

The macros print!, println!, eprint!, eprintln!, write!, writeln! and format! are a special case and implicitly take a reference to any arguments to be formatted.

These macros do not behave as normal functions and macros do for reasons of convenience; the fact that they take references silently is part of that difference.

fn main() {
    let x = 5;
    println!("{}", x);
}

Run it through rustc -Z unstable-options --pretty expanded on the nightly compiler and we can see what println! expands to:

#![feature(prelude_import)]
#[prelude_import]
use std::prelude::v1::*;
#[macro_use]
extern crate std;
fn main() {
    let x = 5;
    {
        ::std::io::_print(::core::fmt::Arguments::new_v1(
            &["", "
"],
            &match (&x,) {
                (arg0,) => [::core::fmt::ArgumentV1::new(
                    arg0,
                    ::core::fmt::Display::fmt,
                )],
            },
        ));
    };
}

Tidied further, it’s this:

use std::{fmt, io};

fn main() {
    let x = 5;
    io::_print(fmt::Arguments::new_v1(
        &["", "
"],
        &[fmt::ArgumentV1::new(&x, fmt::Display::fmt)],
        //                     ^^
    ));
}

Note the &x.

If you write println!("{}", &x), you are then dealing with two levels of references; this has the same result because there is an implementation of std::fmt::Display for &T where T implements Display (shown as impl<'a, T> Display for &'a T where T: Display + ?Sized) which just passes it through. You could just as well write &&&&&&&&&&&&&&&&&&&&&&&x.