问无法将构造函数作为fn使用或强制转换

EN

use std::{iter, ops};

struct Idx(usize);

fn get_inds() -> iter::Zip<iter::Map<ops::RangeFrom<usize>, fn(usize) -> Idx>, ops::RangeFrom<usize>> {
    (0..).map(Idx).zip(0..)
}

error: mismatched types [--explain E0308]
 --> src/main.rs:6:9
6 |>         (0..).map(Idx).zip(0..)
  |>         ^^^^^^^^^^^^^^^^^^^^^^^ expected fn pointer, found fn item
note: expected type `std::iter::Zip<std::iter::Map<std::ops::RangeFrom<usize>, fn(usize) -> Idx>, std::ops::RangeFrom<usize>>`
note:    found type `std::iter::Zip<std::iter::Map<std::ops::RangeFrom<usize>, fn(usize) -> Idx {Idx::{{constructor}}}>, std::ops::RangeFrom<_>>`

fn get_inds() -> iter::Zip<iter::Map<ops::RangeFrom<usize>, fn(usize) -> Idx>, ops::RangeFrom<usize>> {
    (0..).map(Idx as fn(usize) -> Idx).zip(0..)
}

warning: can't cast this type, #[warn(const_err)] on by default
 --> src/main.rs:6:19
6 |>         (0..).map(Idx as fn(usize) -> Idx).zip(0..)
  |>                   ^^^^^^^^^^^^^^^^^^^^^^^