How to write a function generic over pointer type?

Question

In rust-clippy, we have the following function:

fn over<X, F>(left: &[X], right: &[X], mut eq_fn: F) -> bool 
        where F: FnMut(&X, &X) -> bool {
    left.len() == right.len() && left.iter().zip(right).all(|(x, y)| 
        eq_fn(x, y))
}

As it happens, rustc's AST representation uses a lot of syntax::ptr::P<T> pointers. Those dereference to T, and so autoderef implicitly coerces them to &T if we use a closure. If we try to use a plain fn however, we get a type mismatch:

error: type mismatch: the type `fn(&syntax::ast::Expr, &syntax::ast::Expr) -> bool {eq_op::is_exp_equal}` implements the trait `for<'r, 'r> core::ops::FnMut<(&'r syntax::ast::Expr, &'r syntax::ast::Expr)>`, but the trait `for<'r, 'r> core::ops::FnMut<(&'r syntax::ptr::P<syntax::ast::Expr>, &'r syntax::ptr::P<syntax::ast::Expr>)>` is required (expected struct `syntax::ptr::P`, found struct `syntax::ast::Expr`) [E0281]

Can I change the above function to accept both &[&T] and &[P<T>] and automatically coerce P<Expr> into &Expr? If so, how?

Answer 1

Both &T and P<T> implement Deref<Target = T>, so you could use that in your bounds:

use std::ops::Deref;

fn over<X, F, X1, X2>(left: &[X1], right: &[X2], mut eq_fn: F) -> bool 
        where X1: Deref<Target = X>,
              X2: Deref<Target = X>,
              F: FnMut(&X, &X) -> bool {
    left.len() == right.len() && left.iter().zip(right).all(|(x, y)| 
        eq_fn(x, y))
}