基于2元组连接异构列表

Question

考虑基于2元组的类似lisp的异构“列表”：

let a = (); // ()
let b = (1i32, ('2', ("3", (4u64, ())))); // (1i32 '2' "3", 4u64)

我想像这样连接它们：

let a = (1, ('2', ()));
let b = ("3", (4, ()));
let c = concat(a, b); // (1, ('2', ("3", (4, ()))));

我设法以我认为过于复杂的方式这样做了：

impl<Other> Concat<Other> for () {
    type Result = Other;

    fn concat(self, other: Other) -> Self::Result {
        other
    }
}

impl<Head, Tail, Other> Concat<Other> for (Head, Tail)
    where Tail: Concat<Other>
{
    type Result = (Head, <Tail as Concat<Other>>::Result);

    fn concat(self, other: Other) -> Self::Result {
        let (head, tail) = self;
        (head, tail.concat(other))
    }
}

fn main() {
    let a = (1, ('2', ()));
    let b = ("3", (4, ()));
    println!("{:?}", a.concat(b));  // (1, ('2', ("3", (4, ()))));
}

我们可以做得更好吗？如果是这样，怎么样？