Pass impl method to another impl method

help
#1

I'm trying to write a parser.

I want to create a convenience method that will parse a sequence of AST nodes separated by some arbitrary token.

I have this method:

pub fn parse_sequence<T, F> (&mut self, separator: TokenKind, func: F) -> PResult<Vec<T>>
    where
        F: Fn(&mut Parser) -> PResult<T>
    {
        let items = Vec::new();

        items.push(func(self)?);

        while self.tokens.peek().kind == separator {
            self.eat(separator)?;
            items.push(func(self)?);
        }

        Ok(items)
    }

I am calling the function from another method like this:

self.parse_sequence(TokenKind::Comma, Parser::parse_simple_factor)?

With this code I get the following error:

mismatched types
expected type `std::ops::FnOnce<(&mut parse::Parser<'_>,)>`
   found type `for<'r> std::ops::FnOnce<(&'r mut parse::Parser<'_>,)>`rustc(E0308)

This is a tough one. I've tried putting explicit lifetimes all over the parse_sequence function signature, but I always get the same error.

What does this error mean?

Is it possible to pass a method to another method in the same impl?

#2

Have you tried wrapping it in a closure (i.e. |p| p.parse_simple_factor())?

I've found you can't use a fully qualified method (Parser::parse_simple_factor) as a Fn(...). It may be because the method accepts &mut and the compiler can't figure out an appropriate lifetime due to variance... Or something like that.

#3

Thanks for that.

Also just realised that you're the author of that excellent article on geometric constraint solving. It was very informative. When's part two coming? :pleading_face:

#4

Like this?

fn main() {
    let mut s = S;
    s.bar(S::foo);
}

struct S;

impl S {
    fn foo(&mut self) {
        println!("foo!");
    }

    fn bar<F>(&mut self, func: F)
    where
        F: Fn(&mut S),
    {
        func(self);
        println!("bar!");
    }
}

Playground link

#5

Not exactly. It also has to be generic over the return type of the function:

fn bar<F, T>(&mut self, func: F)
where
    F: Fn(&mut S) -> T,
{
    // ...
}
#6

I guess I still don't get it, because this works.

Even if we need the struct to be generic and the return type to match that type, then this works.

#7

I was incorrect earlier. The method should like this:

fn bar<F, T>(&mut self, func: F) -> T
where
    F: Fn(&mut S) -> T,
{
    // ...
   func(self)
}

I just tried that in the playground and it works as well. I'll dig into my code to see what the other differences are.

#8

The error message implies there is a lifetime associated with the struct. Does it have a field which is a reference?

#9

Yeah there is actually. This

#10

It's something about the lifetime of that reference.

Edit: Actually the lifetime introduced by that reference becomes a lifetime on the struct overall, so the function signature needs to reflect that.

This works.

To use your example, just change:

F: Fn(&mut S) -> T,

to

F: Fn(&mut Self) -> T,

That way the compiler knows the lifetime associated with the struct S in the function signature is the same one from the impl block.

#11

Yep. That totally works.

#12

Also, just to show Self isn't magical, you can use

F: Fn(&mut S<'a, T>) -> T,

so long as you no longer elide the lifetime in the impl block, like this:

impl<'a, T: Copy> S<'a, T> {

Playground link

#13

That makes sense.

I couldn't really understand what was happening when Self was used.

#14

I've found using Self is a good habit to have. Since it just means "the thing this impl block is for" it can save a lot of work if you change a type's name or type parameters.