rustownership

Does println! borrow or own the variable?


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.


Solution

  • 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(
                &["", "\n"],
                &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(
            &["", "\n"],
            &[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.


    Early 2023 update:

    • Since mid-2021, the required invocation has been rustc -Zunpretty=expanded rather than rustc -Zunstable-options --pretty=expanded.

    • Since 2023-01-28 or so (https://github.com/rust-lang/rust/pull/106745), format_args! is part of the AST, and so the expansion of println!("{}", x) is ::std::io::_print(format_args!("{0}\n", x));, not exposing the Arguments::new_v1 construction and &x aspects. This is good for various reasons (read #106745’s description), but ruins my clear demonstration here that x was only taken by reference. (This is why I’ve added this as a note at the end rather than updating the answer—since it no longer works.)