我如何解决错误“'Fn'-family traits'类型参数的精确格式可能会发生变化"? [英] How do I solve the error "the precise format of `Fn`-family traits' type parameters is subject to change"?

问题描述

我用 Rust 编写了一个问题解决器，它作为子例程需要调用一个作为黑盒给出的函数(本质上我想给出一个 Fn(f64) -> 类型的参数；f64).

I have written a problem solver in Rust which as a subroutine needs to make calls to a function which is given as a black box (essentially I would like to give an argument of type Fn(f64) -> f64).

本质上我有一个函数定义为 fn solve(f: F) where F : Fn(f64) ->f64 { ... } 这意味着我可以像这样调用 solve :solve(|x| x);

Essentially I have a function defined as fn solve<F>(f: F) where F : Fn(f64) -> f64 { ... } which means that I can call solve like this: solve(|x| x);

我想做的是将一个更复杂的函数传递给求解器，即一个依赖于多个参数等的函数.

What I would like to do is to pass a more complex function to the solver, i.e. a function which depends on multiple parameters etc.

我希望能够将具有合适特征实现的结构传递给求解器.我尝试了以下方法:

I would like to be able to pass a struct with a suitable trait implementation to the solver. I tried the following:

struct Test;
impl Fn<(f64,)> for Test {}

这会产生以下错误:

error: the precise format of `Fn`-family traits' type parameters is subject to change. Use parenthetical notation (Fn(Foo, Bar) -> Baz) instead (see issue #29625)

我还想添加一个特征，其中包括 Fn 特征(不幸的是，我不知道如何定义).这也可能吗?

I would also like to add a trait which includes the Fn trait (which I don't know how to define, unfortunately). Is that possible as well?

澄清一下:我用 C++ 开发已经有一段时间了，C++ 的解决方案是重载 operator()(args).在那种情况下，我可以像函数一样使用 struct 或 class .我希望能够

Just to clarify: I have been developing in C++ for quite a while, the C++ solution would be to overload the operator()(args). In that case I could use a struct or class like a function. I would like to be able to

1. 将函数和结构体作为参数传递给求解器.
2. 有一个简单的方法来调用函数.调用 obj.method(args) 比 obj(args)(在 C++ 中)更复杂.但目前似乎无法实现这种行为.

1. Pass both functions and structs to the solver as arguments.
2. Have an easy way to call the functions. Calling obj.method(args) is more complicated than obj(args) (in C++). But it seems that this behavior is not achievable currently.

推荐答案

直接的答案是完全按照错误消息所说的去做:

The direct answer is to do exactly as the error message says:

用括号代替

也就是说，用Fn(A, B)

真正的问题是您不允许实施Fn* 自己在稳定的 Rust 中的特性家族.

The real problem is that you are not allowed to implement the Fn* family of traits yourself in stable Rust.

您提出的真正问​​题更难确定，因为您没有提供MCVE，因此我们减少了猜测.我会说你应该反过来说；创建一个新特征，为闭包和您的类型实现它:

The real question you are asking is harder to be sure of because you haven't provided a MCVE, so we are reduced to guessing. I'd say you should flip it around the other way; create a new trait, implement it for closures and your type:

trait Solve {
    type Output;
    fn solve(&mut self) -> Self::Output;
}

impl<F, T> Solve for F
where
    F: FnMut() -> T,
{
    type Output = T;

    fn solve(&mut self) -> Self::Output {
        (self)()
    }
}

struct Test;
impl Solve for Test {
    // interesting things
}

fn main() {}

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