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  • Working with Reporting Services Filters – Part 2: The LIKE Operator

    - by smisner
    In the first post of this series, I introduced the use of filters within the report rather than in the query. I included a list of filter operators, and then focused on the use of the IN operator. As I mentioned in the previous post, the use of some of these operators is not obvious, so I'm going to spend some time explaining them as well as describing ways that you can use report filters in Reporting Services in this series of blog posts. Now let's look at the LIKE operator. If you write T-SQL queries, you've undoubtedly used the LIKE operator to produce a query using the % symbol as a wildcard for multiple characters like this: select * from DimProduct where EnglishProductName like '%Silver%' And you know that you can use the _ symbol as a wildcard for a single character like this: select * from DimProduct where EnglishProductName like '_L Mountain Frame - Black, 4_'   So when you encounter the LIKE operator in a Reporting Services filter, you probably expect it to work the same way. But it doesn't. You use the * symbol as a wildcard for multiple characters as shown here: Expression Data Type Operator Value [EnglishProductName] Text Like *Silver* Note that you don’t have to include quotes around the string that you use for comparison. Books Online has an example of using the % symbol as a wildcard for a single character, but I have not been able to successfully use this wildcard. If anyone has a working example, I’d love to see it!

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  • SQL Server Data Type Precedence

    I am executing a simple query/stored procedure from my application against a large table and it's taking a long time to execute. The column I'm using in my WHERE clause is indexed and it's very selective. The search column is not wrapped in a function so that's not the issue. What could be going wrong? Schedule Azure backupsRed Gate’s Cloud Services makes it simple to create and schedule backups of your SQL Azure databases to Azure blob storage or Amazon S3. Try it for free today.

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  • Which to use - "operator new" or "operator new[]" - to allocate a block of raw memory in C++?

    - by sharptooth
    My C++ program needs a block of uninitialized memory. In C I would use malloc() and later free(). In C++ I can either call ::operator new or ::operator new[] and ::operator delete or operator delete[] respectively later. Looks like both ::operator new and ::operator new[] have exactly the same signature and exactly the same behavior. The same for ::operator delete and ::operator delete[]. The only thing I shouldn't do is pairing operator new with operator delete[] and vice versa - undefined behavior. Other than that which pair do I choose and why?

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  • C++ operator new, object versions, and the allocation sizes

    - by mizubasho
    Hi. I have a question about different versions of an object, their sizes, and allocation. The platform is Solaris 8 (and higher). Let's say we have programs A, B, and C that all link to a shared library D. Some class is defined in the library D, let's call it 'classD', and assume the size is 100 bytes. Now, we want to add a few members to classD for the next version of program A, without affecting existing binaries B or C. The new size will be, say, 120 bytes. We want program A to use the new definition of classD (120 bytes), while programs B and C continue to use the old definition of classD (100 bytes). A, B, and C all use the operator "new" to create instances of D. The question is, when does the operator "new" know the amount of memory to allocate? Compile time or run time? One thing I am afraid of is, programs B and C expect classD to be and alloate 100 bytes whereas the new shared library D requires 120 bytes for classD, and this inconsistency may cause memory corruption in programs B and C if I link them with the new library D. In other words, the area for extra 20 bytes that the new classD require may be allocated to some other variables by program B and C. Is this assumption correct? Thanks for your help.

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  • Reflection and Operator Overloads in C#

    - by TenshiNoK
    Here's the deal. I've got a program that will load a given assembly, parse through all Types and their Members and compile a TreeView (very similar to old MSDN site) and then build HTML pages for each node in the TreeView. It basically takes a given assembly and allows the user to create their own MSDN-like library for it for documentation purposes. Here's the problem I've run into: whenever an operator overload is encounted in a defined class, reflection returns that as a "MethodInfo" with the name set to something like "op_Assign" or "op_Equality". I want to be able to capture these and list them properly, but I can't find anything in the MethodInfo object that is returned to accurately identify that I'm looking at an operator. I definitely don't want to just capture everything that starts with "op_", since that will most certainly (at some point) will pick up a method it's not supposed to. I know that other methods and properties that are "special cases" like this one have the "IsSpecialName" property set, but appearantly that's not the case with operators. I've been scouring the 'net and wracking my brain to two days trying to figure this one out, so any help will be greatly appreciated.

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  • Conversion constructor vs. conversion operator: precedence

    - by GRB
    Reading some questions here on SO about conversion operators and constructors got me thinking about the interaction between them, namely when there is an 'ambiguous' call. Consider the following code: class A; class B { public: B(){} B(const A&) //conversion constructor { cout << "called B's conversion constructor" << endl; } }; class A { public: operator B() //conversion operator { cout << "called A's conversion operator" << endl; return B(); } }; int main() { B b = A(); //what should be called here? apparently, A::operator B() return 0; } The above code displays "called A's conversion operator", meaning that the conversion operator is called as opposed to the constructor. If you remove/comment out the operator B() code from A, the compiler will happily switch over to using the constructor instead (with no other changes to the code). My questions are: Since the compiler doesn't consider B b = A(); to be an ambiguous call, there must be some type of precedence at work here. Where exactly is this precedence established? (a reference/quote from the C++ standard would be appreciated) From an object-oriented philosophical standpoint, is this the way the code should behave? Who knows more about how an A object should become a B object, A or B? According to C++, the answer is A -- is there anything in object-oriented practice that suggests this should be the case? To me personally, it would make sense either way, so I'm interested to know how the choice was made. Thanks in advance

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  • C# Nested Property Accessing overloading OR Sequential Operator Overloading

    - by Tim
    Hey, I've been searching around for a solution to a tricky problem we're having with our code base. To start, our code resembles the following: class User { int id; int accountId; Account account { get { return Account.Get(accountId); } } } class Account { int accountId; OnlinePresence Presence { get { return OnlinePresence.Get(accountId); } } public static Account Get(int accountId) { // hits a database and gets back our object. } } class OnlinePresence { int accountId; bool isOnline; public static OnlinePresence Get(int accountId) { // hits a database and gets back our object. } } What we're often doing in our code is trying to access the account Presence of a user by doing var presence = user.Account.Presence; The problem with this is that this is actually making two requests to the database. One to get the Account object, and then one to get the Presence object. We could easily knock this down to one request if we did the following : var presence = UserPresence.Get(user.id); This works, but sort of requires developers to have an understanding of the UserPresence class/methods that would be nice to eliminate. I've thought of a couple of cool ways to be able to handle this problem, and was wondering if anyone knows if these are possible, if there are other ways of handling this, or if we just need to think more as we're coding and do the UserPresence.Get instead of using properties. Overload nested accessors. It would be cool if inside the User class I could write some sort of "extension" that would say "any time a User object's Account property's Presence object is being accessed, do this instead". Overload the . operator with knowledge of what comes after. If I could somehow overload the . operator only in situations where the object on the right is also being "dotted" it would be great. Both of these seem like things that could be handled at compile time, but perhaps I'm missing something (would reflection make this difficult?). Am I looking at things completely incorrectly? Is there a way of enforcing this that removes the burden from the user of the business logic? Thanks! Tim

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  • PHP ternary operator

    - by thecoshman
    ($DAO->get_num_rows() == 1) ? echo("is") : echo("are"); This dose not seem to be working for me,I get an error "Unexpected T_ECHO" I have tried it with out the brackets around the conditional. Am I just not able to use a ternary operator in this way? The $DAO-get_num_rows() returns an integer value. Thanks

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  • redefine __and__ operator

    - by wiso
    Why I can't redefine the __and__ operator? class Cut(object): def __init__(self, cut): self.cut = cut def __and__(self, other): return Cut("(" + self.cut + ") && (" + other.cut + ")") a = Cut("a>0") b = cut("b>0") c = a and b print c.cut() I want (a>0) && (b>0), but I got b, that the usual behaviour of and

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  • Regex AND operator

    - by user366735
    Based on this answer http://stackoverflow.com/questions/469913/regular-expressions-is-there-an-and-operator I tried the following on http://regexpal.com/ but was unable to get it to work. What am missing? Does javascript not support it? Regex: (?=foo)(?=baz) String: foo,bar,baz

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  • subscript operator on pointers

    - by Lodle
    If i have a pointer to an object that has an overloaded subscript operator ( [] ) why cant i do this: MyClass *a = new MyClass(); a[1]; but have to do this instead: MyClass *a = new MyClass(); (*a)[1];

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  • Using operator+ without leaking memory?

    - by xokmzxoo
    So the code in question is this: const String String::operator+ (const String& rhs) { String tmp; tmp.Set(this->mString); tmp.Append(rhs.mString); return tmp; } This of course places the String on the stack and it gets removed and returns garbage. And placing it on the heap would leak memory. So how should I do this?

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  • how to call operator () in c++

    - by anish
    in c++ i have following code class Foobar{ public: Foobar * operator()(){ return new Foobar; } My quesion is how to call the (); if i do Foobar foo() the constructor gets called i am confused about behaviour of () can some explain me

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  • typedef and operator overloading in C

    - by jocapco
    Suppose I typedef an integer or integer array or any known type: typedef int int2 Then I overload operator * for int2 pairs, now if I initialize variables a and b as int. Then will my * between a and b be the overloaded * ? How do I achieve overloading an int and yet also use * for int the way they are. Should I create a new type?

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  • Where namespace does operator<< (stream) go to?

    - by aaa
    If I have have some overloaded ostream operators, defined for library local objects, is its okay for them to go to std namespace? If I do not declare them in std namespace, then I must use using ns:: operator <<. As a possible follow-up question, are there any operators which should go to standard or global namespace?

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  • Are there supposed to be more restrictions on operator->* overloads?

    - by Potatoswatter
    I was perusing section 13.5 after refuting the notion that built-in operators do not participate in overload resolution, and noticed that there is no section on operator->*. It is just a generic binary operator. Its brethren, operator->, operator*, and operator[], are all required to be non-static member functions. This precludes definition of a free function overload to an operator commonly used to obtain a reference from an object. But the uncommon operator->* is left out. In particular, operator[] has many similarities. It is binary (they missed a golden opportunity to make it n-ary), and it accepts some kind of container on the left and some kind of locator on the right. Its special-rules section, 13.5.5, doesn't seem to have any actual effect except to outlaw free functions. (And that restriction even precludes support for commutativity!) So, for example, this is perfectly legal (in C++0x, remove obvious stuff to translate to C++03): #include <utility> #include <iostream> #include <type_traits> using namespace std; template< class F, class S > typename common_type< F,S >::type operator->*( pair<F,S> const &l, bool r ) { return r? l.second : l.first; } template< class T > T & operator->*( pair<T,T> &l, bool r ) { return r? l.second : l.first; } template< class T > T & operator->*( bool l, pair<T,T> &r ) { return l? r.second : r.first; } int main() { auto x = make_pair( 1, 2.3 ); cerr << x->*false << " " << x->*4 << endl; auto y = make_pair( 5, 6 ); y->*(0) = 7; y->*0->*y = 8; // evaluates to 7->*y = y.second cerr << y.first << " " << y.second << endl; } I can certainly imagine myself giving into temp[la]tation. For example, scaled indexes for vector: v->*matrix_width[5] = x; Did the standards committee forget to prevent this, was it considered too ugly to bother, or are there real-world use cases?

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