parser: Remove empty multiline string parts earlier

Makes parsing more consistent and is a super minor optimisation

Co-authored-by: Robert Hensing <[email protected]>

src/libexpr/parser-state.hh

@@ -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;

tests/functional/lang/parse-okay-ind-string.exp

@@ -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") ])