rust - 为什么将闭包传递给接受函数指针的函数不起作用？

Question

在second edition of The Rust Programming Language (强调我的):

Function pointers implement all three of the closure traits (Fn, FnMut, and FnOnce), so you can always pass a function pointer as an argument for a function that expects a closure. It’s best to write functions using a generic type and one of the closure traits so your functions can accept either functions or closures.

将闭包传递给接受函数指针作为参数的函数不会编译:

fn main() {
    let a = String::from("abc");

    let x = || println!("{}", a);

    fn wrap(c: fn() -> ()) -> () {
        c()
    }

    wrap(x);
}

error[E0308]: mismatched types
  --> src/main.rs:10:10
   |
10 |     wrap(x);
   |          ^ expected fn pointer, found closure
   |
   = note: expected type `fn()`
              found type `[closure@src/main.rs:4:13: 4:33 a:_]`

为什么这不起作用？

Answer 1

闭包不是函数。

您可以将一个函数传递给一个需要闭包的函数，但没有理由反之亦然。

如果您希望能够将闭包和函数都作为参数传递，只需为闭包做好准备即可。

例如:

let a = String::from("abc");

let x = || println!("x {}", a);

fn y() {
    println!("y")
}

fn wrap(c: impl Fn()) {
    c()
}

wrap(x); // pass a closure
wrap(y); // pass a function