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parser: Remove empty multiline string parts earlier
Makes parsing more consistent and is a super minor optimisation Co-authored-by: Robert Hensing <roberth@users.noreply.github.com>
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2 changed files with 13 additions and 2 deletions
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@ -291,12 +291,23 @@ inline Expr * ParserState::stripIndentation(const PosIdx pos,
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s2 = std::string(s2, 0, p + 1);
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}
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es2->emplace_back(i->first, new ExprString(std::move(s2)));
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// Ignore empty strings for a minor optimisation and AST simplification
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if (s2 != "") {
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es2->emplace_back(i->first, new ExprString(std::move(s2)));
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}
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};
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for (; i != es.end(); ++i, --n) {
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std::visit(overloaded { trimExpr, trimString }, i->second);
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}
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// If there is nothing at all, return the empty string directly.
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// This also ensures that equivalent empty strings result in the same ast, which is helpful when testing formatters.
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if (es2->size() == 0) {
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auto *const result = new ExprString("");
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delete es2;
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return result;
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}
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/* If this is a single string, then don't do a concatenation. */
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if (es2->size() == 1 && dynamic_cast<ExprString *>((*es2)[0].second)) {
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auto *const result = (*es2)[0].second;
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@ -1 +1 @@
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(let string = "str"; in [ (/some/path) ((/some/path)) (("" + /some/path)) ((/some/path + "\n end")) (string) ((string)) (("" + string)) ((string + "\n end")) ("") ("") ("end") ])
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(let string = "str"; in [ (/some/path) ((/some/path)) ((/some/path)) ((/some/path + "\n end")) (string) ((string)) ((string)) ((string + "\n end")) ("") ("") ("end") ])
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