我有一个通用枚举,其中只有一个变体使用通用类型。
[将ParseResult<T>
“转换”为ParseResult<U>
时,编译器也迫使我破坏了非通用部分。非通用部分将立即重新组装,并以完全相同的方式返回。
没有重复代码是否有更优雅的解决方案?
#[derive(PartialEq, Debug)]
enum ParseResult<'a, T> {
Success(T, &'a str),
Failure(&'static str),
}
fn char<'a>(c: char) -> impl Fn(&'a str) -> ParseResult<'a, char> {
move |input| match input.chars().next() {
Some(candidate) if c == candidate => ParseResult::Success(candidate, &input[1..]),
Some(_) => ParseResult::Failure("chars didn't match"),
None => ParseResult::Failure("unexpected end of stream"),
}
}
fn or<'a, T>(
mut lhs: impl FnMut(&'a str) -> ParseResult<T>,
mut rhs: impl FnMut(&'a str) -> ParseResult<T>,
) -> impl FnMut(&'a str) -> ParseResult<T> {
move |input| match lhs(input) {
ParseResult::Failure(_) => rhs(input),
ok => ok,
}
}
fn and<'a, T, U>(
mut lhs: impl FnMut(&'a str) -> ParseResult<T>,
mut rhs: impl FnMut(&'a str) -> ParseResult<U>,
) -> impl FnMut(&'a str) -> ParseResult<(T, U)> {
move |input| match lhs(input) {
ParseResult::Success(left, after) => match rhs(after) {
ParseResult::Success(right, after) => ParseResult::Success((left, right), after),
// Only explicit works:
ParseResult::Failure(why) => ParseResult::Failure(why),
},
// Same as line as above, same goes here.
ParseResult::Failure(why) => ParseResult::Failure(why),
}
}
不是最后两条线::Failure
无效的事情:
bad => bad,
预期元组,找到类型参数U
bad => bad as ParseResult<(T, U)>,
bad @ ParseResult::Failure(why) => bad,
预期元组,找到类型参数U
但是非通用部分会立即以完全相同的方式重新组装并返回。没有代码重复,没有更聪明/更优雅的解决方案吗?