rust - 为什么对实现 Fn 的类型的引用不被识别为可调用？

Question

即使 &T 被定义为实现 Fn 特性，编译器在将其作为可调用对象调用时也会拒绝它:

trait Trait {
    fn act(self);
}

//passes
fn test_ref_input_as_trait<'a, T>(t: &'a T)
where
    &'a T: Trait,
{
    t.act();
}

//fails
fn test_ref_input_as_fntrait<'a, T>(t: &'a T)
where
    &'a T: Fn(),
{
    t();
}

//passes
fn test_input_as_fntrait<T>(t: T)
where
    T: Fn(),
{
    t();
}

编译器拒绝第二个函数的定义:

error[E0618]: expected function, found `&'a T`
  --> src/lib.rs:18:5
   |
14 | fn test_ref_input_as_fntrait<'a, T>(t: &'a T)
   |                                     - `&'a T` defined here
...
18 |     t();
   |     ^^^ not a function

对于 nightly (1.32)，错误消息替换为:

error[E0618]: expected function, found `&'a T`
  --> src/lib.rs:18:5
   |
14 | fn test_ref_input_as_fntrait<'a, T>(t: &'a T)
   |                                     - `&'a T` defined here
...
18 |     t();
   |     ^--
   |     |
   |     call expression requires function

也许我遗漏了什么，但为什么编译器接受定义但不允许调用它？是我这边的句法缺陷导致它理解别的东西吗？

Answer 1

有一个open issue (#42736)对这个。然而，docs for Fn状态:

for any type F that implements Fn, &F implements Fn, too.

这意味着以下工作:

fn test_ref_input_as_fntrait<'a, T>(t: &'a T)
where
    T: Fn(),
{
    t();
}