我有以下带有匹配表达式的函数:
fn handle_event<'e>(&mut self, event: Event<'e>) -> Event<'e> {
match (&event, &self.current_lang) {
(Event::Start(Tag::CodeBlock(CodeBlockKind::Fenced(lang))), _) => {
self.start_fenced_code_block(&lang)
}
(Event::End(Tag::CodeBlock(CodeBlockKind::Fenced(_))), _) => {
self.end_fenced_code_block()
}
(Event::Text(text), Some(lang)) => self.code_html(&text, &lang),
_ => event,
}
}
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然而,由于深层嵌套的枚举,前两个手臂感觉像是失控了。所以我做了一些宏:
macro_rules! fenced_code_block_start {
($lang:pat_param) => {
Event::Start(Tag::CodeBlock(CodeBlockKind::Fenced($lang)))
};
}
macro_rules! fenced_code_block_end {
() => {
Event::End(Tag::CodeBlock(CodeBlockKind::Fenced(_)))
};
}
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现在我有了,恕我直言,更清洁的:
match (&event, &self.current_lang) {
(fenced_code_block_start!(lang), _) => self.start_fenced_code_block(&lang),
(fenced_code_block_end!(), _) => self.end_fenced_code_block(),
(Event::Text(text), Some(lang)) => self.code_html(&text, &lang),
_ => event,
}
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但是,我想知道是否有更好的方法来做到这一点。陷入宏总是让我停顿。我可以在这里使用 Rust 的其他一些功能,而不是那么沉重的锤子吗?
如果不丢失清晰度,您可以直接导入并使用枚举变体:
fn handle_event<'e>(&mut self, event: Event<'e>) -> Event<'e> {
use Event::*;
use Tag::*;
use CodeBlockKind::*;
match (&event, &self.current_lang) {
(Start(CodeBlock(Fenced(lang))), _) => self.start_fenced_code_block(&lang),
(End(CodeBlock(Fenced(_))), _) => self.end_fenced_code_block(),
(Text(text), Some(lang)) => self.code_html(&text, &lang),
_ => event,
}
}
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