传达类型参数的 Rust 生命周期

Question

I'm working on a simple complex number example, and trying to implement ref-value/value-ref operations as follows:

use std::ops::*;

#[derive(Clone, PartialEq)]
pub struct Complex<T: Sized + Clone> {
    pub re: T,
    pub im: T,
}

// Ref-Ref Multiplication
impl<'a, 'b, T: Sized + Clone> Mul<&'b Complex<T>> for &'a Complex<T>
where
    T: Add<T, Output = T>,
    T: Sub<T, Output = T>,
    &'a T: Add<&'b T, Output = T>,
    &'a T: Mul<&'b T, Output = T>,
    &'a T: Sub<&'b T, Output = T>,
{
    type Output = Complex<T>;
    fn mul(self, rhs: &'b Complex<T>) -> Complex<T> {
        panic!("// Details irrelevant")
    }
}

// Ref-Value Multiplication
impl<'a, 'b, T: Sized + Clone> Mul<Complex<T>> for &'a Complex<T>
where
    T: 'static,
    T: Add<T, Output = T>,
    T: Sub<T, Output = T>,
    &'a T: Add<&'b T, Output = T>,
    &'a T: Mul<&'b T, Output = T>,
    &'a T: Sub<&'b T, Output = T>,
{
    type Output = Complex<T>;
    fn mul(self, rhs: Complex<T>) -> Complex<T> {
        let t = &rhs;
        self.mul(t)
    }
}

The ref-ref implementation works, and from what I understand it it takes in two references of differing lifetimes, and returns a complex value-type. The ref-value part is where I'm having an issue; When I compile, the error is that rhs doesn't live long enough. I believe I know why this is already, and that is that T could hold a reference (either direct or indirectly) to rhs when the value is returned, thus rhs goes out of scope, but T could hold a reference to it still.

My question is how to communicate that T will not hold some reference to rhs in some shape or form.

Some notes on things that I've tried so far or looked at:

1. Changed the lifetime specification on either Mul implementation.
2. Tried lifetime-inheritence, but this specifieds a reference held by T will live at least as long as T , so I think I need something more in the lines of "at most."
3. Looked at other implementations; Either does not implement the case, or just uses clone to bypass the issue.

Answer 1

As suggested by Peter Hall in the comments, the easiest solution is to derive Copy for your complex type, and implement the operations for values. For the ref-ref implementations and the ref-val implementations, you can then simply dereference the references and use the val-val implementation.

If you want to make the approach you started work, you need higher-rank trait bounds:

use std::ops::*;

#[derive(Clone, PartialEq)]
pub struct Complex<T: Clone> {
    pub re: T,
    pub im: T,
}

// Ref-Ref Multiplication
impl<'a, 'b, T: Clone> Mul<&'b Complex<T>> for &'a Complex<T>
where
    T: Add<T, Output = T>,
    T: Sub<T, Output = T>,
    &'a T: Add<&'b T, Output = T>,
    &'a T: Mul<&'b T, Output = T>,
    &'a T: Sub<&'b T, Output = T>,
{
    type Output = Complex<T>;
    fn mul(self, rhs: &'b Complex<T>) -> Complex<T> {
        Complex {
            re: &self.re * &rhs.re - &self.im * &rhs.im,
            im: &self.re * &rhs.im + &self.im * &rhs.re,
        }
    }
}

// Ref-Value Multiplication
impl<'a, T: Clone> Mul<Complex<T>> for &'a Complex<T>
where
    T: Add<T, Output = T>,
    T: Sub<T, Output = T>,
    &'a T: for<'b> Add<&'b T, Output = T>,
    &'a T: for<'b> Mul<&'b T, Output = T>,
    &'a T: for<'b> Sub<&'b T, Output = T>,
{
    type Output = Complex<T>;
    fn mul(self, rhs: Complex<T>) -> Complex<T> {
        let t = &rhs;
        self.mul(t)
    }
}

In your version, the lifetime 'b in the ref-value implementation is chosen by the user of the trait. Since the user could use any lifetime for 'b , rhs would need static lifetime for your code to be valid. What you want instead is that *'a T satisfies the given trait bounds for any given lifetime 'b , which is exactly what HRTBs are for.

An alternative, less repetitive way of writing the trait bounds for the second implementation is this:

impl<'a, T: Clone> Mul<Complex<T>> for &'a Complex<T>
where
    Self: for<'b> Mul<&'b Complex<T>, Output = Complex<T>>,
{
    type Output = Complex<T>;
    fn mul(self, rhs: Complex<T>) -> Complex<T> {
        self.mul(&rhs)
    }
}

Answer 2

The built-in numeric types implement these permutations using a macro . Doing it by hand, I'd start with the case where you are multiplying two values, rather than any references, and make sure that your Complex struct is Copy :

impl<T: Copy> Mul<Complex<T>> for Complex<T>
where
    T: Add<T, Output = T>,
    T: Sub<T, Output = T>,
    T: Mul<T, Output = T>,
{
    type Output = Complex<T>;
    fn mul(self, rhs: Complex<T>) -> Complex<T> {
        unimplemented!()
    }
}

Note that you don't need any constraints on &T - you can't return references to T anyway, so you are going to have to copy them, which is why I've specified T: Copy .

The rest of the implementations are now straightforward and can delegate to the simplest case:

impl<'a, T: Copy> Mul<Complex<T>> for &'a Complex<T>
where
    T: Add<T, Output = T>,
    T: Sub<T, Output = T>,
    T: Mul<T, Output = T>,
{
    type Output = Complex<T>;
    fn mul(self, rhs: Complex<T>) -> Complex<T> {
        (*self).mul(rhs)
    }
}

impl<'a, T: Copy> Mul<&'a Complex<T>> for Complex<T>
where
    T: Add<T, Output = T>,
    T: Sub<T, Output = T>,
    T: Mul<T, Output = T>,
{
    type Output = Complex<T>;
    fn mul(self, rhs: &'a Complex<T>) -> Complex<T> {
        self.mul(*rhs)
    }
}

impl<'a, 'b, T: Copy> Mul<&'a Complex<T>> for &'b Complex<T>
where
    T: Add<T, Output = T>,
    T: Sub<T, Output = T>,
    T: Mul<T, Output = T>,
{
    type Output = Complex<T>;
    fn mul(self, rhs: &'a Complex<T>) -> Complex<T> {
        (*self).mul(*rhs)
    }
}