Help with mut lifetimes

help
1

I am trying to parse XML using quick_xml into some struct with zero allocation. I think this is MRE of my problem:

enum Results {
    One,
    Two,
}

struct Reader<R> {
    reader: R,
}

impl Reader<&[u8]> {
    fn read_into<'a>(&mut self, buf: &'a mut Vec<u8>) -> Results {
        todo!()
    }

    fn read_text(&mut self) -> &str {
        ""
    }
}

struct Doc<'a> {
    id: &'a str,
}

fn read_root<'a>(rd: &'a mut Reader<&'a [u8]>, doc: &'a mut Doc<'a>) {
    let mut buf: Vec<u8> = Vec::new();

    loop {
        match rd.read_into(&mut buf) {
            Results::One => {
                read_child_1(rd, doc);
            }
            Results::Two => {
                read_child_2(rd);
            }
        }

        buf.clear();
    }
}

fn read_child_1<'a>(rd: &'a mut Reader<&'a [u8]>, doc: &'a mut Doc<'a>) {
    doc.id = rd.read_text();
}

fn read_child_2(rd: &mut Reader<&[u8]>) {}

The compiler complains about multiple mutable borrowing when "read_child_1" is not commented out, but doesn't if it is commented out. The difference is second argument. What am I missing here?

2

The problem is because of your definition of read_child_1 which conflicts with the definition of the Doc struct:

fn read_child_1<'a>(rd: &'a mut Reader<&'a [u8]>, doc: &'a mut Doc<'a>) {
    doc.id = rd.read_text();
}

Here you are saying that you are borrowing Doc for a lifetime 'a, but this lifetime comes from the time when you borrow it:

match rd.read_into(&mut buf) {
            Results::One => {
                read_child_1(rd, doc);
            }

Thus effectively making the 'a lifetime that you defined inside read_child_1 to equal the 'a lifetime defined at read_root.

I would start by not making Doc hold a lifetime.

3

Can you explain? Where exactly should I do that?

4
5

Yes, that I know. Changing everything to owned types fixes everything. But this is not what I want. I don't want to allocate. I already have an owned XML string, I don't want to create a second copy of it.
Also, if I could process everything in the "parent" function that would work too. But my XML is very complex and I need to call multiple "child" functions to process it.
If zero allocation is impossible in my case, it is fine; I just want to make sure it is, before giving up on it.

6

As pointed out above, &'a mut Struct<'a> should never be used in Rust. So the simplest solution is to reduce lifetimes somewhere to have less trouble.

If doc insists to have a lifetime, the lifetime annotation should be

fn read_root<'id>(rd: &'id mut Reader<&[u8]>, doc: &mut Doc<'id>) {
    let mut buf: Vec<u8> = Vec::new();

    // loop {
        ...
    // }
}
fn read_child_1<'id>(rd: &'id mut Reader<&'_ [u8]>, doc: &'_ mut Doc<'id>) { doc.id = rd.read_text(); }

and for simplicity, successful code without loop: Rust Playground

Then, based on the code above, one pattern with loops can be Rust Playground

fn read_root(rd: &mut Reader<&[u8]>) {
    let mut buf: Vec<u8> = Vec::new();

    loop {
        match rd.read_into(&mut buf) {
            Results::One => {
                let mut doc = Doc { id: /* temprary lifetime 'id */ };
                read_child_1(/*&'id mut*/rd, &mut doc); // &mut Reader<&[u8]> reborrows with 'id
            }
            Results::Two => {
                read_child_2(rd);
            }
        }

        buf.clear();
    }
}
fn read_child_1<'id>(rd: &'id mut Reader<&'_ [u8]>, doc: &'_ mut Doc<'id>) {
    doc.id = rd.read_text();
}

I guess this won't be your expected code. But this may be the best result with &mut.

You can go a little further along & road, since it's Copy, and &'id can be everywhere (at the cost of interior/shared mutability): Rust Playground

struct Reader<R> {
    reader: Mutex<R>,
}
fn read_root<'id>(rd: &'id Reader<&[u8]>, doc: &mut Doc<'id>) {
    let mut buf: Vec<u8> = Vec::new();

    loop {
        match rd.read_into(&mut buf) {
            Results::One => {
                read_child_1(rd, doc);
            }
            Results::Two => {
                read_child_2(rd);
            }
        }

        buf.clear();
    }
}

fn read_child_1<'id>(rd: &'id Reader<&'_ [u8]>, doc: &'_ mut Doc<'id>) {
    doc.id = rd.read_text();
}
4 Likes
7

Overlapping with @vague’s answer (they were faster than me), but here is a slightly different working version I got in the playground: Rust Playground

enum Results {
    One,
    Two,
}

struct Reader<R> {
    reader: R,
}

impl<'a> Reader<&'a [u8]> {
    fn read_into(&mut self, _buf: &mut Vec<u8>) -> Results {
        todo!()
    }

    fn read_text(&mut self) -> &'a str {
        ""
    }
}

struct Doc<'a> {
    id: &'a str,
}

fn read_root<'a>(rd: &mut Reader<&'a [u8]>, doc: &mut Doc<'a>) {
    let mut buf: Vec<u8> = Vec::new();

    loop {
        let results = rd.read_into(&mut buf); 

        match results {
            Results::One => {
                read_child_1(rd, doc);
            }
            Results::Two => {
                read_child_2(rd);
            }
        }

        buf.clear();
    }
}

fn read_child_1<'a>(rd: &mut Reader<&'a [u8]>, doc: &mut Doc<'a>) {
    doc.id = rd.read_text();
}

fn read_child_2(_rd: &mut Reader<&[u8]>) {}

Reiterating this: beware of &'a mut Struct<'a>

3 Likes
8

Thank you; looks promising. Let me try this with real code.
Beautiful! I used this signature for both parent and child functions and everything works great now:

fn <'a>(rd: &mut Reader<&'a [u8]>, doc: &mut Doc<'a>)