예제 #1
0
        public Expression IsOfMethod(IsOfMethod method, LiteralExpression[] arguments)
        {
            Debug.Assert(arguments[1].LiteralType == LiteralType.String);

            var type = LiteralUtil.GetCompatibleType((string)arguments[1].Value);

            return(new LiteralExpression(type.IsInstanceOfType(arguments[0].Value), LiteralType.Boolean));
        }
 public virtual TResult IsOfMethod(IsOfMethod method, Expression[] arguments)
 {
     throw new QueryNotSupportException();
 }