> This is because of higher ranked lifetime bounds...
No, it's because the function you're returning is of type FnMut(&B) -> C, but your F function is of type FnMut(B) -> G. It expects an owned value of B, but you only have a borrowed reference to a B. Just make F of type FnMut(&B) -> G and you no longer have to dereference and no longer need the restriction that B: Copy. Higher-ranked lifetime values don't have anything to do with it.
You think I didn't try that one before? It doesn't work in MY case:
error[E0281]: type mismatch: the type `fn(_) -> _ {tool::second::<_>}` implements the trait `std::ops::FnMut<(_,)>`, but the trait `for<'r> std::ops::FnMut<(&'r _,)>` is required (expected concrete lifetime, found bound lifetime parameter
--> src\lib.rs:50:17
|
50 | .filter(apply(second, i))
| ^^^^^
|
= note: required by `apply`
error[E0271]: type mismatch resolving `for<'r> <fn(_) -> _ {tool::second::<_>} as std::ops::FnOnce<(&'r _,)>>::Output == _`
--> src\lib.rs:50:17
|
50 | .filter(apply(second, i))
| ^^^^^ expected bound lifetime parameter , found concrete lifetime
|
= note: concrete lifetime that was found is lifetime '_#11r
= note: required by `apply`
error: aborting due to 2 previous errors
error: Could not compile `fizzbuzz`.
No, it's because the function you're returning is of type FnMut(&B) -> C, but your F function is of type FnMut(B) -> G. It expects an owned value of B, but you only have a borrowed reference to a B. Just make F of type FnMut(&B) -> G and you no longer have to dereference and no longer need the restriction that B: Copy. Higher-ranked lifetime values don't have anything to do with it.
Here's the new version: