Do not fail to evaluate an expression if some input is missing

My keyboard layout does not have Alt_R but ISO_Level3_Shift. As a
consequence any control expression containing Alt_R fails to evaluate
completely and is unusable. This modification replace the missing term
of the expression by a dummy expression which always evaluate to
0. This way, the keybinding can work even if some keys are not
available.

1 files changed, 31 insertions, 1 deletions

diff --git a/Source/Core/InputCommon/ControllerInterface/ExpressionParser.cpp b/Source/Core/InputCommon/ControllerInterface/ExpressionParser.cpp
index 8c9060e69c..b49d4b8dce 100644
--- a/Source/Core/InputCommon/ControllerInterface/ExpressionParser.cpp
+++ b/Source/Core/InputCommon/ControllerInterface/ExpressionParser.cpp
@@ -219,6 +219,33 @@ public:
 	virtual operator std::string() { return ""; }
 };
 
+class DummyExpression : public ExpressionNode
+{
+public:
+	std::string name;
+
+	DummyExpression(const std::string& name_) : name(name_) {}
+
+	ControlState GetValue() override
+	{
+		return 0.0;
+	}
+
+	void SetValue(ControlState value) override
+	{
+	}
+
+	int CountNumControls() override
+	{
+		return 0;
+	}
+
+	operator std::string() override
+	{
+		return "`" + name + "`";
+	}
+};
+
 class ControlExpression : public ExpressionNode
 {
 public:
@@ -416,7 +443,10 @@ private:
 			{
 				Device::Control *control = finder.FindControl(tok.qualifier);
 				if (control == nullptr)
-					return EXPRESSION_PARSE_NO_DEVICE;
+				{
+					*expr_out = new DummyExpression(tok.qualifier);
+					return EXPRESSION_PARSE_SUCCESS;
+				}
 
 				*expr_out = new ControlExpression(tok.qualifier, control);
 				return EXPRESSION_PARSE_SUCCESS;