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parser: Remove empty multiline string parts earlier

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Makes parsing more consistent and is a super minor optimisation

Co-authored-by: Robert Hensing <roberth@users.noreply.github.com>
This commit is contained in:
Silvan Mosberger 2024-07-17 02:43:42 +02:00 committed by Silvan Mosberger
parent 9fae50ed4b
commit 0c91bb97e5
2 changed files with 13 additions and 2 deletions
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13
src/libexpr/parser-state.hh
View file

@ -291,12 +291,23 @@ inline Expr * ParserState::stripIndentation(const PosIdx pos,
s2 = std::string(s2, 0, p + 1);
}
es2->emplace_back(i->first, new ExprString(std::move(s2)));
// Ignore empty strings for a minor optimisation and AST simplification
if (s2 != "") {
es2->emplace_back(i->first, new ExprString(std::move(s2)));
}
};
for (; i != es.end(); ++i, --n) {
std::visit(overloaded { trimExpr, trimString }, i->second);
}
// If there is nothing at all, return the empty string directly.
// This also ensures that equivalent empty strings result in the same ast, which is helpful when testing formatters.
if (es2->size() == 0) {
auto *const result = new ExprString("");
delete es2;
return result;
}
/* If this is a single string, then don't do a concatenation. */
if (es2->size() == 1 && dynamic_cast<ExprString *>((*es2)[0].second)) {
auto *const result = (*es2)[0].second;

2
tests/functional/lang/parse-okay-ind-string.exp
View file

@ -1 +1 @@
(let string = "str"; in [ (/some/path) ((/some/path)) (("" + /some/path)) ((/some/path + "\n end")) (string) ((string)) (("" + string)) ((string + "\n end")) ("") ("") ("end") ])
(let string = "str"; in [ (/some/path) ((/some/path)) ((/some/path)) ((/some/path + "\n end")) (string) ((string)) ((string)) ((string + "\n end")) ("") ("") ("end") ])