Understanding Rust Error E0700

Rust Error E0700 Explained

Rust Error E0700 occurs when a function with an impl Trait return type captures lifetime parameters that do not appear within the impl Trait itself. This can lead to ambiguity and potential issues with the borrow checker.

Example of Erroneous Code

#![allow(unused)]
fn main() {
  use std::cell::Cell;

  trait Trait<'a> { }

  impl<'a, 'b> Trait<'b> for Cell<&'a u32> { }

  fn foo<'x, 'y>(x: Cell<&'x u32>) -> impl Trait<'y>
  where 'x: 'y
  {
      x
  }
}

In the example above, function foo returns a value of type Cell<&'x u32>, which references the lifetime 'x. However, the return type is declared as impl Trait<'y>. This means that foo returns a type that implements Trait<'y> and only captures data referencing lifetime 'y. Since we are referencing data with lifetime 'x, this function generates Rust Error E0700.

Solution for Rust Error E0700

To fix the Rust Error E0700, reference the correct lifetime in the return type. In this case, changing the return type to impl Trait<'y> + 'x will work:

#![allow(unused)]
fn main() {
  use std::cell::Cell;

  trait Trait<'a> { }

  impl<'a, 'b> Trait<'b> for Cell<&'a u32> { }

  fn foo<'x, 'y>(x: Cell<&'x u32>) -> impl Trait<'y> + 'x
  where 'x: 'y
  {
      x
  }
}

By adjusting the return type as shown, the function foo now correctly captures the lifetime 'x, and the Rust Error E0700 is resolved.