HLSL: Fix a grammar error related to constructors in parenthetical expressions
diff --git a/hlsl/hlslGrammar.cpp b/hlsl/hlslGrammar.cpp
index e3e5df7..11b6dc1 100755
--- a/hlsl/hlslGrammar.cpp
+++ b/hlsl/hlslGrammar.cpp
@@ -1718,27 +1718,29 @@
if (acceptTokenClass(EHTokLeftParen)) {
TType castType;
if (acceptType(castType)) {
- if (! acceptTokenClass(EHTokRightParen)) {
- expected(")");
- return false;
+ if (acceptTokenClass(EHTokRightParen)) {
+ // We've matched "(type)" now, get the expression to cast
+ TSourceLoc loc = token.loc;
+ if (! acceptUnaryExpression(node))
+ return false;
+
+ // Hook it up like a constructor
+ TFunction* constructorFunction = parseContext.handleConstructorCall(loc, castType);
+ if (constructorFunction == nullptr) {
+ expected("type that can be constructed");
+ return false;
+ }
+ TIntermTyped* arguments = nullptr;
+ parseContext.handleFunctionArgument(constructorFunction, arguments, node);
+ node = parseContext.handleFunctionCall(loc, constructorFunction, arguments);
+
+ return true;
+ } else {
+ // This could be a parenthesized constructor, ala (int(3)), and we just accepted
+ // the '(int' part. We must back up twice.
+ recedeToken();
+ recedeToken();
}
-
- // We've matched "(type)" now, get the expression to cast
- TSourceLoc loc = token.loc;
- if (! acceptUnaryExpression(node))
- return false;
-
- // Hook it up like a constructor
- TFunction* constructorFunction = parseContext.handleConstructorCall(loc, castType);
- if (constructorFunction == nullptr) {
- expected("type that can be constructed");
- return false;
- }
- TIntermTyped* arguments = nullptr;
- parseContext.handleFunctionArgument(constructorFunction, arguments, node);
- node = parseContext.handleFunctionCall(loc, constructorFunction, arguments);
-
- return true;
} else {
// This isn't a type cast, but it still started "(", so if it is a
// unary expression, it can only be a postfix_expression, so try that.