我正在尝试在Rust中实现类型级别的乘法。
添加已经有效,但我遇到了“临时”类型变量的问题。
代码:
use std::marker::PhantomData;
//Trait for the type level naturals
trait Nat {}
impl Nat for Zero {}
impl<T: Nat> Nat for Succ<T> {}
//Zero and successor types
struct Zero;
struct Succ<T: Nat>(PhantomData<T>);
//Type level addition
trait Add<B,C>
where Self: Nat,
B: Nat,
C: Nat
{}
impl<B: Nat> Add<B,B> for Zero {}
impl<A: Nat,B: Nat,C: Nat> Add<B,C> for Succ<A>
where A: Add<Succ<B>,C>
{}
fn add<A: Nat, B: Nat, C: Nat>(
a: PhantomData<A>,
b: PhantomData<B>)
-> PhantomData<C>
where A: Add<B,C> { PhantomData }
//Type level multiplication
trait Mult<B,C>
where Self: Nat,
B: Nat,
C: Nat,
{}
impl<B: Nat> Mult<B,Zero> for Zero {}
//ERROR HERE: "unconstrained type parameter 'C'"
//impl<A: Nat, B: Nat,C: Nat, D: Nat> Mult<B,D> for Succ<A>
// where A: Mult<B,C>,
// B: Add<C,D>
// {}
fn main() {
let x: PhantomData<Succ<Succ<Zero>>> = PhantomData;
let y: PhantomData<Succ<Zero>> = PhantomData;
//uncomment ': i32' in the next line to see infered type
let z /*: i32*/ = add(x,y);
}
发布的代码编译得很好并且添加工作。 如果我取消注释ERROR HERE部分,我会收到以下错误消息:
error[E0207]: the type parameter `C` is not constrained by the impl trait, self type, or predicates
--> src/main.rs:40:21
|
40 | impl<A: Nat, B: Nat,C: Nat, D: Nat> Mult<B,D> for Succ<A>
| ^ unconstrained type parameter
error: aborting due to previous error
error: Could not compile `4_18_generics`.
To learn more, run the command again with --verbose.
有没有办法使用这种“临时/中间”类型参数?
是否可以以任何其他方式进行乘法(我目前没想到)?
一般不可能吗?
该语言的未来版本是否可行?
答案 0 :(得分:5)
我认为你滥用了泛型,这是你问题的根源。
Rust中的泛型具有输入和输出:
<>
直觉是,对于给定的输入集,每个输出都选择一种类型。
在你的情况下,我们必须重新修改这个特征:
trait Add<Rhs: Nat>: Nat {
type Result: Nat;
}
特质的定义说:
Add
要求Self
为Nat
Add
是针对右侧参数实现的,该参数必须为Nat
Add
的特定实施具有Result
类型,必须为Nat
现在我们可以实现它:
impl<T: Nat> Add<T> for Zero {
type Result = T;
}
impl<A: Nat, B: Nat> Add<B> for Succ<A>
where A: Add<Succ<B>>
{
type Result = < A as Add<Succ<B>> >::Result;
}
请注意,功能完全没必要,“A + B”的结果是:
<A as Add<B>>::Result
现在,进行乘法运算:
trait Mul<Rhs: Nat>: Nat {
type Result: Nat;
}
impl<T: Nat> Mul<T> for Zero {
type Result = Zero;
}
// The first attempt does not work, but I'm keeping it here as
// it is good for learning purpose.
//
// impl<A: Nat, B: Nat> Mul<B> for Succ<A>
// where A: Mul<B> + Add< <A as Mul<B>>::Result >
// {
// type Result = <A as Add< <A as Mul<B>>::Result >>::Result;
// }
//
// Think:
// 1. Why exactly does it not work?
// 2. What exactly is going on here?
// 3. How would you multiply numbers in terms of addition?
// 4. m * n = m + m + m ... (n times)? Or: n + n + .. (m times)?
//
// Answering these questions will help learning the intricacies of
// Rust's traits/type-system and how they work.
impl<A: Nat, B: Nat> Mul<B> for Succ<A>
where
A: Mul<B>,
B: Add<<A as Mul<B>>::Result>,
{
type Result = <B as Add<<A as Mul<B>>::Result>>::Result;
}
现在编译:
fn main() {
type One = Succ<Zero>;
type Two = <One as Add<One>>::Result;
type Four = <Two as Mul<Two>>::Result;
}
请注意,此类Peano算法具有相当令人讨厌的限制,特别是在递归的深度。你的加法是O(N),你的乘法是O(N ^ 2),......
如果您对更有效的陈述和计算感兴趣,我建议您查看最先进的typenum箱。