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Fix logic for generating unique symbol map

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Previously, this didn't correctly handle the case where *multiple* 
symbols were all simply-aliased to the same *other* symbol.

Refs #500
This commit is contained in:
Max Brunsfeld 2020-01-27 11:20:48 -08:00
parent 9ffcb16392
commit 8dd68c360a
1 changed files with 7 additions and 11 deletions
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18
cli/src/generate/render.rs
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@ -325,22 +325,18 @@ impl Generator {
add_line!(self, "static TSSymbol ts_symbol_map[] = {{");
indent!(self);
for symbol in &self.parse_table.symbols {
// There can be multiple symbols in the grammar that have the same name and kind,
// due to simple aliases. When that happens, ensure that they map to the same
// public-facing symbol. If one of the symbols is not aliased, choose that one
// to be the public-facing symbol. Otherwise, pick the symbol with the lowest
// numeric value.
let mut mapping = symbol;
// If this symbol has a simple alias, then check if its alias has the same
// name and kind (e.g. named vs anonymous) as some other symbol in the grammar.
// If so, add an entry to the symbol map that deduplicates these two symbols,
// so that only one of them will ever be returned via the public API.
if let Some(alias) = self.simple_aliases.get(symbol) {
let kind = alias.kind();
for other_symbol in &self.parse_table.symbols {
if other_symbol == symbol {
continue;
}
if let Some(other_alias) = self.simple_aliases.get(other_symbol) {
if other_symbol < symbol && other_alias == alias {
if other_symbol < mapping && other_alias == alias {
mapping = other_symbol;
break;
}
} else if self.metadata_for_symbol(*other_symbol) == (&alias.value, kind) {
mapping = other_symbol;
@ -353,7 +349,7 @@ impl Generator {
self,
"[{}] = {},",
self.symbol_ids[&symbol],
self.symbol_ids[&mapping],
self.symbol_ids[mapping],
);
}