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GDScript: Fix error message for Lua-style dictionary

4.1
AleryBerry 2023-03-20 21:44:27 +07:00
parent a7d0e18a31
commit 0550f3682b
1 changed files with 2 additions and 2 deletions
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4
modules/gdscript/gdscript_parser.cpp
Unescape Escape View File

@ -2731,12 +2731,12 @@ GDScriptParser::ExpressionNode *GDScriptParser::parse_dictionary(ExpressionNode
switch (dictionary->style) {
case DictionaryNode::LUA_TABLE:
if (key != nullptr && key->type != Node::IDENTIFIER && key->type != Node::LITERAL) {
push_error("Expected identifier or string as LUA-style dictionary key.");
push_error(R"(Expected identifier or string as Lua-style dictionary key (e.g "{ key = value }").)");
advance();
break;
}
if (key != nullptr && key->type == Node::LITERAL && static_cast<LiteralNode *>(key)->value.get_type() != Variant::STRING) {
push_error("Expected identifier or string as LUA-style dictionary key.");
push_error(R"(Expected identifier or string as Lua-style dictionary key (e.g "{ key = value }").)");
advance();
break;
}