How to return a reference in Rust Closure

Question

I have the following rust code cannot be compiled.

struct Person {
    name : String,
    age : u8,
}


fn main() {
    let p = Person{ name: "Nobody".to_string(), age : 24};

    let age = |p : &Person| p.age;
    let name = |p : &Person | &p.name;

    println! ("name={}, age={}" , name(&p), age(&p));
}

And the compiler gave the following error message.

   Compiling playground v0.0.1 (/playground)
error[E0495]: cannot infer an appropriate lifetime for borrow expression due to conflicting requirements
  --> src/main.rs:11:31
   |
11 |     let name = |p : &Person | &p.name;
   |                               ^^^^^^^
   |
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the body at 11:16...
  --> src/main.rs:11:16
   |
11 |     let name = |p : &Person | &p.name;
   |                ^^^^^^^^^^^^^^^^^^^^^^
note: ...so that reference does not outlive borrowed content
  --> src/main.rs:11:31
   |
11 |     let name = |p : &Person | &p.name;
   |                               ^^^^^^^
note: but, the lifetime must be valid for the expression at 2:29...
  --> src/main.rs:13:5
   |
13 |     println! ("name={}, age={}" , name(&p), age(&p));
   |     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
note: ...so type `(&&std::string::String, &u8)` of expression is valid during the expression
  --> src/main.rs:13:5
   |
13 |     println! ("name={}, age={}" , name(&p), age(&p));
   |     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
   = note: this error originates in a macro outside of the current crate (in Nightly builds, run with -Z external-macro-backtrace for more info)

error: aborting due to previous error

I tried to add the lifetime for name closure.

let name<'a> = |p : &'a Person | -> &'a String { &'a p.name };

but still got the compiler error

   Compiling playground v0.0.1 (/playground)
error: expected one of `:`, `;`, `=`, `@`, or `|`, found `<`
  --> src/main.rs:12:13
   |
12 |     let name<'a> = |p : &'a Person | -> &'a String { &'a p.name };
   |             ^ expected one of `:`, `;`, `=`, `@`, or `|`

error: aborting due to previous error

Just want to know how to write the correct code.

Answer 1

For me it is more important to understand the source of the problem that find a workaround, so step by step. let's start from something that works:

struct Person {
    name: String,
    age: u8,
}

fn get_name<'a>(person: &'a Person) -> &'a str {
    &person.name
}

fn main() {
    let p = Person {
        name: "Nobody".to_string(),
        age: 24,
    };

    let age = |p: &Person| p.age;
    let name = get_name;

    println!("name={}, age={}", name(&p), age(&p));
}

There is no issues when using a function instead of a clousure. In this case, the compiler is able to check that lifetime requirements are ok.

But when trying to use a closure for name :

let name = |p : &Person | &p.name;

You get the cannot infer an appropriate lifetime error.

Why?

A closure captures its environment: some opaque struct has to be created by the compiler and such struct has to be callable.

I'm non fully aware of the internal details, but something along these lines would be created when desugaring your closure:

struct OpaqueType<'a> {
    // a PhantomData because you don't capure nothing
    // just to make explicit that struct lifetime bind to environment
    // if you would had captured some integer:
    // captured_int: &'a i32,
    captured_int: PhantomData<&'a i32>,

}

impl<'a> OpaqueType<'a> {
    fn call<'b>(&'b self, person: &'a Person) -> &'a str {
        &person.name
    }
}

And looking at call it is apparent that when a clusure argument is a reference there are two unrelated lifetimes at play.

Finally the answer: how to return a reference

Also note that in your case, not declaring the argument type and using the helper function get_name , works:

// let name = |p| &p.name; // does not work, not enough info to infer p type 
let name = |p| get_name(p);

My guess is that in such case the compiler, following some inference path, is able to desugar in a way that lifetimes are bounded as expected.

Answer 2

One other solution is to give an explicit type to your closure. Unfortunately, you can't use its actual type, but you can cast it to a function pointer.

Remember that the issue is that the compiler isn't able to correctly deduce that the lifetime of the output is tied to the lifetime of the input (it could be an instance of this bug , but I'm not at all sure). We can fix that by making the lifetimes explicit.

struct Person {
    name: String,
    age: u8,
}

fn main() {
    let p = Person {
        name: "Nobody".to_string(),
        age: 24,
    };

    let age = |p: &Person| p.age;
    // Our only changes are right here.
    let name: for<'a> fn(&'a Person) -> &'a String = |p: &Person| &p.name;

    println!("name={}, age={}", name(&p), age(&p));
}

(playground)

In fact, it's possible to be slightly less explicit than this. The compiler is fine figuring out the types of the input and output. It's just the lifetimes it has trouble with. So replacing that line with let name: for<'a> fn(&'a _) -> &'a _ = |p: &Person| &p.name; let name: for<'a> fn(&'a _) -> &'a _ = |p: &Person| &p.name; also works (playground) .