fix: idx1 of variable expression without initializer

Fix Idx1 of VariableExpression so that it points to the character right after the name literal if the expression does not have an initializer.

ast/node.go

@@ -483,7 +483,7 @@ func (ve *VariableExpression) Idx0() file.Idx {
 // Idx1 implements Node.
 func (ve *VariableExpression) Idx1() file.Idx {
 	if ve.Initializer == nil {
-		return file.Idx(int(ve.Idx) + len(ve.Name) + 1)
+		return file.Idx(int(ve.Idx) + len(ve.Name))
 	}
 	return ve.Initializer.Idx1()
 }

parser/parser_test.go

@@ -1086,6 +1086,20 @@ func TestPosition(t *testing.T) {
 		node = block.List[0].(*ast.ExpressionStatement).Expression.(*ast.UnaryExpression)
 		is(node.Idx0(), 14)
 		is(parser.slice(node.Idx0(), node.Idx1()), "xyz++")
+
+		parser = newParser("", "(function(){ var abc, xyz = 1; })", 1, nil)
+		program, err = parser.parse()
+		is(err, nil)
+		block = program.Body[0].(*ast.ExpressionStatement).Expression.(*ast.FunctionLiteral).Body.(*ast.BlockStatement)
+		node = block.List[0].(*ast.VariableStatement)
+		is(node.Idx0(), 14)
+		is(parser.slice(node.Idx0(), node.Idx1()), "var abc, xyz = 1")
+		node = block.List[0].(*ast.VariableStatement).List[0].(*ast.VariableExpression)
+		is(node.Idx0(), 18)
+		is(parser.slice(node.Idx0(), node.Idx1()), "abc")
+		node = block.List[0].(*ast.VariableStatement).List[1].(*ast.VariableExpression)
+		is(node.Idx0(), 23)
+		is(parser.slice(node.Idx0(), node.Idx1()), "xyz = 1")
 	})
 }