rust - Rust:通过工厂方法创建结构时出现问题

Question

采取这个最小的repro(playground)，在这里我尝试将带有闭包的结构作为字段，并定义一些其他工厂方法，以闭包的形式提供自定义逻辑。

#![allow(non_snake_case)]

struct FunStruct<T>
where
    T: Fn(i32) -> i32
{
    pub p: T,
}

fn Create<T>() -> FunStruct<T>
where
    T: Fn(i32) -> i32
{
    let p : T = |val| -> i32 {
        return val;
    };
    
    let a = FunStruct {
        p,
    };
    
    return a;
}

fn main() {
    let a = Create();
    
    println!("{}", (a.p)(5));
}

上面的代码段引发以下错误。

error[E0308]: mismatched types
  --> src/main.rs:14:17
   |
10 |   fn Create<T>() -> FunStruct<T>
   |             - this type parameter
...
14 |       let p : T = |val| -> i32 {
   |  _____________-___^
   | |             |
   | |             expected due to this
15 | |         return val;
16 | |     };
   | |_____^ expected type parameter `T`, found closure
   |
   = note: expected type parameter `T`
                     found closure `[closure@src/main.rs:14:17: 16:6]`

error[E0282]: type annotations needed for `FunStruct<T>`
  --> src/main.rs:26:13
   |
26 |     let a = Create();
   |         -   ^^^^^^ cannot infer type for type parameter `T` declared on the function `Create`
   |         |
   |         consider giving `a` the explicit type `FunStruct<T>`, where the type parameter `T` is specified
   |
   = note: type must be known at this point

error: aborting due to 2 previous errors

Some errors have detailed explanations: E0282, E0308.
For more information about an error, try `rustc --explain E0282`.

但是，这似乎很好

struct FunStruct<T>
where
    T: Fn(i32) -> i32
{
    pub p: T,
}

fn main() {
    let p = |val| -> i32 {
        return val;
    };
    
    let a = FunStruct {
        p
    };
    
    println!("{}", (a.p)(5));
}

某种程度的抽象似乎毁了它。

Answer 1

通用参数意味着它适用于任何类型-但是，Create中的闭包不适用于任何闭包类型，它会创建具有一个特定类型的闭包。你可以这样做:

fn Create() -> FunStruct<impl Fn(i32) -> i32>{
    let p/*: some anonymous, unique type */ = |val| -> i32 {
        return val;
    };
    
    let a = FunStruct {
        p,
    };
    
    return a;
}

返回位置中的 impl Trait表示“我返回实现此特征的某种类型，虽然我不会告诉您它实际上是什么”。相反，您的原始函数说“给定一个函数的任何类型，我都可以返回它”-这显然没有意义。