rust - 如何将泛型函数转换为带有引用参数的函数指针？

Question

pub fn remove_file<P: AsRef<Path>>(path: P) -> Result<()>;

我正在努力将 std::fs::remove_file 转换为函数指针:
Playground

use std::{io, fs, path::Path};

fn main() {
    let _:  fn(&Path) -> io::Result<()> = &fs::remove_file;

    // let _: &dyn FnOnce(&Path) -> io::Result<()> = &fs::remove_file;
}

错误如下:

error[E0308]: mismatched types
 --> src/main.rs:4:43
  |
4 |     let _:  fn(&Path) -> io::Result<()> = &fs::remove_file;
  |             ---------------------------   ^^^^^^^^^^^^^^^^ expected fn pointer, found reference
  |             |
  |             expected due to this
  |
  = note: expected fn pointer `for<'r> fn(&'r std::path::Path) -> std::result::Result<(), std::io::Error>`
              found reference `&fn(_) -> std::result::Result<(), std::io::Error> {std::fs::remove_file::<_>}`

问题可能与 for<'r>高阶生存期要求有关，但我不知道如何解决。
如何使用特征对象？以下内容也不编译:

let _: &dyn FnOnce(&Path) -> io::Result<()> = &fs::remove_file;

我知道我可以围绕 fs::remove_file制作包装函数，但我想避免这种情况，并让 fs::remove_file成为函数指针或特征对象本身。
这不是此 Function pointers in Rust using constrained generics的副本，因为我想使用 std::fs::remove_file作为类型参数获取指向 &Path具体实例化的函数指针，但不具有泛型函数指针类型。
我尝试了以下操作，但也不起作用:

use std::{io, fs, path::Path};

fn main() {
    let _:  fn(&Path) -> io::Result<()> = &fs::remove_file::<&Path>;
}

Answer 1

I'd like to get a function pointer to the concrete instantiation of std::fs::remove_file with &Path as a type parameter, but not to have a generic function pointer type.

这是问题的核心。一个具体的实例将被编译。例如:

let _ : fn(&'static Path) -> io::Result<()> = fs::remove_file::<&'static Path>;
let _ : fn(&'r Path) -> io::Result<()> = fs::remove_file::<&'r Path>;
let _ : fn(&'long Path) -> io::Result<()> = fs::remove_file::<&'short Path>;

您尝试做的是从 fs::remove_file类型强制:

∀ P: AsRef<Path>, P → io::Result<()>

类型:

∀ 'r, &'r Path → io::Result<()>

在整个生命周期中，函数指针类型仍然是通用的。我不认为无法执行这种强制性是有原因的，但是是 Rust has some issues in this area。
Rust理解的最直接的解决方法是使用闭包:

let _ : fn(&Path) -> io::Result<()> = |p| fs::remove_file(p);