rust - 将泛型函数作为参数传递

Question

我希望能够将通用函数传递给另一个函数(在本例中为闭包)，而不会丢失所传递函数的“通用性”。由于这是一个相当复杂的陈述，这里有一个例子:

use std::fmt::Debug;

fn test<F, I: Debug>(gen: F) where F: Fn(fn(I) -> I) -> I {
    fn input<I: Debug>(x: I) -> I {
        x
    }
    
    println!("{:?}", gen(input));
}

fn main() {
    test(|input| {
        input(10);
        input(10.0)
    });
}

这不会编译，因为 input 的值是类型推断的，不再是通用的。

完整错误:

<anon>:14:15: 14:19 error: mismatched types:
 expected `_`,
    found `_`
(expected integral variable,
    found floating-point variable) [E0308]
<anon>:14         input(10.0)
                        ^~~~

在 rust 中这样的事情可能吗？

编辑:

根据给出的解决方案，我使用以下方法解决了类似的问题:

#![feature(unboxed_closures)]
#![feature(fn_traits)]

use std::ops::Fn;
use std::ops::Add;
use std::ops::FnMut;

use std::fmt::Debug;

struct Builder;

impl Builder {
    pub fn build<A: Add<B>, B: Add<A>>(&self) -> fn(A, B) -> <A as std::ops::Add<B>>::Output {
        fn c<A: Add<B>, B: Add<A>>(a: A, b: B) -> <A as std::ops::Add<B>>::Output {
            a + b
        }
        
        return c;
    }
}

impl<A: Add<B>, B: Add<A>> Fn<(A, B)> for Builder {
    extern "rust-call" fn call(&self, args: (A, B)) -> <A as std::ops::Add<B>>::Output {
        let (a1, a2) = args;
        self.build()(a1, a2)
    }
}

impl<A: Add<B>, B: Add<A>> FnMut<(A, B)> for Builder {
    extern "rust-call" fn call_mut(&mut self, args: (A, B)) -> <A as std::ops::Add<B>>::Output {
        let (a1, a2) = args;
        self.build()(a1, a2)
    }
}

impl<A: Add<B>, B: Add<A>> FnOnce<(A, B)> for Builder {
    type Output = <A as std::ops::Add<B>>::Output;
    extern "rust-call" fn call_once(self, args: (A, B)) -> <A as std::ops::Add<B>>::Output {
        let (a1, a2) = args;
        self.build()(a1, a2)
    }
}

fn test<F, I: Debug>(gen: F) where F: Fn(Builder) -> I {
    let b = Builder;
    println!("{:?}", gen(b));
}

fn main() {
    test(|builder| {
        builder(10, 10);
        builder(10.1, 10.0)
    });
}

Answer 1

如前所述，不幸的是调用在调用点是单态的，所以你不能传递泛型函数，你只能传递泛型函数的单态版本。

然而，您可以传递的是函数构建器:

use std::fmt::Debug;

struct Builder;

impl Builder {
    fn build<I: Debug>(&self) -> fn(I) -> I {
        fn input<I: Debug>(x: I) -> I { x }
        input
    }
}

fn test<F, T: Debug>(gen: F)
    where F: Fn(Builder) -> T
{
    let builder = Builder;
    println!("{:?}", gen(builder));
}

fn main() {
    test(|builder| {
        builder.build()(10);
        builder.build()(10.0)
    });
}

Builder 能够根据需要生成 input 的实例。