doc: Remove a wrong information in core::ptr::eq<T>()

[ShoyuVanilla]

It says that PartialEq for &T implementation calls core::ptr::eq<T> but this is not true as &T does not implement PartialEq unless T implements PartialEq; the following code is not compiled.

struct Foo;

fn main() {
    let a = Foo;
    let b = Foo;
    assert!(&a == &b);
}

error[E0369]: binary operation `==` cannot be applied to type `&Foo`
 --> src/main.rs:6:16
  |
6 |     assert!(&a == &b);
  |             -- ^^ -- &Foo
  |             |
  |             &Foo
  |
note: an implementation of `PartialEq` might be missing for `Foo`
 --> src/main.rs:1:1
  |
1 | struct Foo;
  | ^^^^^^^^^^ must implement `PartialEq`
help: consider annotating `Foo` with `#[derive(PartialEq)]`
  |
1 + #[derive(PartialEq)]
2 | struct Foo;
  |

For more information about this error, try `rustc --explain E0369`.

And even if it implements PartialEq, it does not actually call core::ptr::eq<T> for the reference comparision.

#[derive(PartialEq)]
struct Foo;

fn main() {
    let a = Foo;
    let b = Foo;
    assert!(&a == &b);
    assert!(!std::ptr::eq::<Foo>(&a, &b));
}

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[compiler-errors]

I think you may be misinterpreting what the parentheses are trying to say. Specifically:

rather than comparing the values they point to (which is what the PartialEq for &T implementation does)

This is saying that core::ptr::eq::<T> does not compare the values that the references point to, and is saying that PartialEq for &T does compare the values that the references point to.

The wording on master is not incorrect as far as I can tell.

[ShoyuVanilla]

I think you may be misinterpreting what the parentheses are trying to say. Specifically:

rather than comparing the values they point to (which is what the PartialEq for &T implementation does)

This is saying that core::ptr::eq::<T> does not compare the values that the references point to, and is saying that PartialEq for &T does compare the values that the references point to.

The wording on master is not incorrect as far as I can tell.

Oh, what a silly misreading 😅 thanks!