Is it possible to have a struct which contains a reference to a value which has a shorter lifetime than the struct?

Question

Here is a simplified version of what I want to archive:

struct Foo<\'a> {
    boo: Option<&\'a mut String>,
}

fn main() {
    let mut foo =          


        

Answer 1

This is obviously completely safe

What is obvious to humans isn't always obvious to the compiler; sometimes the compiler isn't as smart as humans (but it's way more vigilant!).

In this case, your original code compiles when non-lexical lifetimes are enabled:

#![feature(nll)]

struct Foo<'a> {
    boo: Option<&'a mut String>,
}

fn main() {
    let mut foo = Foo { boo: None };
    {
        let mut string = "Hello".to_string();
        foo.boo = Some(&mut string);
        foo.boo.unwrap().push_str(", I am foo!");
        foo.boo = None;
    } // string goes out of scope. foo does not reference string anymore

} // foo goes out of scope

This is only because foo is never used once it would be invalid (after string goes out of scope), not because you set the value to None. Trying to print out the value after the innermost scope would still result in an error.

Is it possible to have a struct which contains a reference to a value which has a shorter lifetime than the struct?

The purpose of Rust's borrowing system is to ensure that things holding references do not live longer than the referred-to item.

After non-lexical lifetimes

Maybe, so long as you don't make use of the reference after it is no longer valid. This works, for example:

#![feature(nll)]

struct Foo<'a> {
    boo: Option<&'a mut String>,
}

fn main() {
    let mut foo = Foo { boo: None };
    // This lives less than `foo`
    let mut string1 = "Hello".to_string();
    foo.boo = Some(&mut string1); 
    // This lives less than both `foo` and `string1`!
    let mut string2 = "Goodbye".to_string();
    foo.boo = Some(&mut string2); 
}

Before non-lexical lifetimes

No. The borrow checker is not smart enough to tell that you cannot / don't use the reference after it would be invalid. It's overly conservative.

In this case, you are running into the fact that lifetimes are represented as part of the type. Said another way, the generic lifetime parameter 'a has been "filled in" with a concrete lifetime value covering the lines where string is alive. However, the lifetime of foo is longer than those lines, thus you get an error.

The compiler does not look at what actions your code takes; once it has seen that you parameterize it with that specific lifetime, that's what it is.

---

The usual fix I would reach for is to split the type into two parts, those that need the reference and those that don't:

struct FooCore {
    size: i32,
}

struct Foo<'a> {
    core: FooCore, 
    boo: &'a mut String,
}

fn main() {
    let core = FooCore { size: 42 };
    let core = {
        let mut string = "Hello".to_string();
        let foo = Foo { core, boo: &mut string };
        foo.boo.push_str(", I am foo!");
        foo.core        
    }; // string goes out of scope. foo does not reference string anymore

} // foo goes out of scope

Note how this removes the need for the Option — your types now tell you if the string is present or not.

An alternate solution would be to map the whole type when setting the string. In this case, we consume the whole variable and change the type by changing the lifetime:

struct Foo<'a> {
    boo: Option<&'a mut String>,
}

impl<'a> Foo<'a> {
    fn set<'b>(self, boo: &'b mut String) -> Foo<'b> {
        Foo { boo: Some(boo) }
    }

    fn unset(self) -> Foo<'static> {
        Foo { boo: None }
    }
}

fn main() {
    let foo = Foo { boo: None };
    let foo = {
        let mut string = "Hello".to_string();
        let mut foo = foo.set(&mut string);
        foo.boo.as_mut().unwrap().push_str(", I am foo!");
        foo.unset()
    }; // string goes out of scope. foo does not reference string anymore

} // foo goes out of scope