"为实现另一个特征的所有具体类型一般地实现一个特征"

Question

I'm pretty new to rust, so this might be a stupid question with an obvious answer, but I don't know the answer yet. I've just read the book and then decided to implement some algorithms I'm familiar with, to get a feeling for the language. That, however, took me down a dark path to a side project that went a little deeper into the language than I had planned for...

def_offset!(Offset);
def_idx!(
    Idx
    with offset Offset
    with sub [
        Vec<T>[Idx] => T,
        [T][Idx] => T
    ]
);

Answer 1

Looking at the playground example, it's not possible to do what you want. This type of impl can't work:

impl<X, RHS> std::ops::Add<RHS> for X
where
    X: MyAdd<RHS>,
{
    type Output = X::Res;
    fn add(self, rhs: RHS) -> Self::Output {
        self.my_add(rhs)
    }
}

Answer 2

I fought it a bit more, and I think I might have a solution. Since I can't implement a trait for general types, but only types in my own crate, that is what I have to do. I didn't want to, because I wanted to use the types to define different operators, but I figured that there would be a way around that.

// Any of the wrapped types should have this.
pub trait TypeInfo: Copy {
    type WrappedType: Copy + num::PrimInt;
}