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[parser] Handle the result of passing a subparser's result to the next parser in the stack

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Eryn Wells 2018-08-25 07:58:49 -07:00
parent 580a71c997
commit 683c9df243
1 changed files with 6 additions and 5 deletions
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11
parser/src/lib.rs
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@ -97,16 +97,17 @@ impl<T> Iterator for Parser<T> where T: Iterator<Item=LexerResult> {
Some(NodeParseResult::Complete{ obj }) => { Some(NodeParseResult::Complete{ obj }) => {
println!("{:?} completed with {:?}", self.parsers.last().unwrap(), obj); println!("{:?} completed with {:?}", self.parsers.last().unwrap(), obj);
self.pop_parser(); self.pop_parser();
if self.parsers.len() == 0 && input_lex.is_none() { if self.parsers.len() == 0 {
// We are done. println!("we are done parsing");
println!("we are done");
out = Some(Ok(obj)); out = Some(Ok(obj));
break; break;
} else { } else {
let prev_parser = self.parsers.last_mut().unwrap(); let prev_parser = self.parsers.last_mut().unwrap();
prev_parser.subparser_completed(obj); println!("passing completed obj {:?} to previous parser {:?}", obj, prev_parser);
// TODO: Handle the result from above. result = Some(prev_parser.subparser_completed(obj));
continue;
} }
// TODO: This is dead code now... Is this correct?
println!("parsers {:?}", self.parsers); println!("parsers {:?}", self.parsers);
self.next_lex() self.next_lex()
}, },