Why isn't `std::mem::drop` exactly the same as the closure |_|() in higher-ranked trait bounds?

The core of the issue is that drop is not a single function, but rather a parameterized set of functions that each drop some particular type. To satisfy a higher-ranked trait bound (hereafter hrtb), you'd need a single function that can simultaneously take references to a type with any given lifetime.

We'll use drop as our typical example of a generic function, but all this applies more generally too. Here's the code for reference: fn drop<T>(_: T) {}.

Conceptually, drop is not a single function, but rather one function for every possible type T. Any particular instance of drop takes only arguments of a single type. This is called monomorphization. If a different T is used with drop, a different version of drop is compiled. That's why you can't pass a generic function as an argument and use that function in full generality (see this question)

On the other hand, a function like fn pass(x: &i32) -> &i32 {x} satisfies the hrtb for<'a> Fn(&'a i32) -> &'a i32. Unlike drop, we have a single function that simultaneously satisfies Fn(&'a i32) -> &'a i32 for every lifetime 'a. This is reflected in how pass can be used.

fn pass(x: &i32) -> &i32 {

fn two_uses<F>(f: F)
    for<'a> F: Fn(&'a i32) -> &'a i32, // By the way, this can simply be written
                                       // F: Fn(&i32) -> &i32 due to lifetime elision rules.
                                       // That applies to your original example too.
        // x has some lifetime 'a
        let x = &22;
        println!("{}", f(x));
        // 'a ends around here
        // y has some lifetime 'b
        let y = &23;
        println!("{}", f(y));
        // 'b ends around here
    // 'a and 'b are unrelated since they have no overlap

fn main() {


In the example, the lifetimes 'a and 'b have no relation to each other: neither completely encompasses the other. So there isn't some kind of subtyping thing going on here. A single instance of pass is really being used with two different, unrelated lifetimes.

This is why drop doesn't satisfy for<'a> FnOnce(&'a T). Any particular instance of drop can only cover one lifetime (ignoring subtyping). If we passed drop into two_uses from the example above (with slight signature changes and assuming the compiler let us), it would have to choose some particular lifetime 'a and the instance of drop in the scope of two_uses would be Fn(&'a i32) for some concrete lifetime 'a. Since the function would only apply to single lifetime 'a, it wouldn't be possible to use it with two unrelated lifetimes.

So why does the toilet closure get a hrtb? When inferring the type for a closure, if the expected type hints that a higher-ranked trait bound is needed, the compiler will try to make one fit. In this case, it succeeds.

Issue #41078 is closely related to this and in particular, eddyb's comment here gives essentially the explanation above (though in the context of closures, rather than ordinary functions). The issue itself doesn't address the present problem though. It instead addresses what happens if you assign the toilet closure to a variable before using it (try it out!).

It's possible that the situation will change in the future, but it would require a pretty big change in how generic functions are monomorphized.

In short, both lines should fail. But since one step in old way of handling hrtb lifetimes, namely the leak check, currently has some soundness issue, rustc ends up (incorrectly) accepting one and leaving the other with a pretty bad error message.

If you disable the leak check with rustc +nightly -Zno-leak-check, you'll be able to see a more sensible error message:

error[E0308]: mismatched types
  --> src/main.rs:10:5
10 |     foo(drop, "drop");
   |     ^^^ one type is more general than the other
   = note: expected type `std::ops::FnOnce<(&'a &str,)>`
              found type `std::ops::FnOnce<(&&str,)>`

My interpretation of this error is that the &x in the body of the foo function only has a scope lifetime confined to the said body, so f(&x) also has the same scope lifetime which can't possibly satisfy the for<'a> universal quantification required by the trait bound.

The question you present here is almost identical to issue #57642, which also has two contrasting parts.

The new way to process hrtb lifetimes is by using so-called universes. Niko has a WIP to tackle the leak check with universes. Under this new regime, both parts of issue #57642 linked above is said to all fail with far more clear diagnoses. I suppose the compiler should be able to handle your example code correctly by then, too.