参数化结构的运算符依赖于另一个运算符

Question

我正在用 Rust 编写 LoG 过滤器，我想使用 | 作为元素宽乘法运算符 (a_{i,j} * b_{i,j}) 的运算符，但编译器抱怨 Output 结果。它说 self[(i, j)] * out[(i, j)] 不等于 Mul<N>::Output。

impl<N> BitOr<Matrix<N>> for Matrix<N> where N: Mul<N> {
    type Output = Matrix<N>;

    fn bitor(self, other: Matrix<N>) -> Matrix<N> {
        if self.width() != other.width() ||
            self.height() != other.height() {
                panic!("Matrices need to have equal dimensions");
            }

        let mut out = Matrix::new(self.width(), self.height());

        for i in 0..(self.width()) {
            for j in 0..(self.height()) {
                out[(i, j)] = self[(i, j)] * out[(i, j)];
            }
        }

        out
    }
}

有什么方法可以根据 Mul<N>::Output 类型设置输出吗？

Answer 1

我想这应该可行：

impl<N> BitOr<Matrix<N>> for Matrix<N> where N: Mul<N> {
    type Output = Matrix<<N as Mul<N>>::Output>;

    fn bitor(self, other: Matrix<N>) -> Matrix<<N as Mul<N>>::Output> {
        if self.width() != other.width() ||
            self.height() != other.height() {
                panic!("Matrices need to have equal dimensions");
            }

        let mut out = Matrix::new(self.width(), self.height());

        for i in 0..(self.width()) {
            for j in 0..(self.height()) {
                out[(i, j)] = self[(i, j)] * out[(i, j)];
            }
        }

        out
    }
}

Answer 2

你没有提供可运行的小例子，所以我自己做了一个。这有效：

use std::ops::{Mul,BitOr};

#[derive(Copy,Show)]
struct Matrix<N>(N, N);

impl<N> BitOr<Matrix<N>> for Matrix<N> where N: Mul<N, Output=N> {
    type Output = Matrix<N>;

    fn bitor(self, other: Matrix<N>) -> Matrix<N> {
        Matrix(self.0 * other.0, self.1 * other.1)
    }
}

fn main() {
    let a = Matrix(-1,-1);
    let b = Matrix(2,3);
    let c = a | b;

    println!("{:?}", c)
}

我必须做的主要事情是 N: Mul<N, Output=N>，它指定 N 必须乘以另一个 N 和将导致在另一个 N.

参数化结构的运算符依赖于另一个运算符

Operator for parametrized struct is dependant on another operator

operator-overloading

multiplication

rust