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  • Showplan Operator of the Week - Compute Scalar

    The third part of Fabiano's mission to describe the major Showplan Operators used by SQL Server's Query Optimiser continues with the 'Compute Scalar' operator. Fabiano shows how a tweak to SQL to avoid a 'Compute Scalar' step can improve its performance.

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  • Showplan Operator of the Week - Lazy Spool

    Continuing to illuminate the depths of SQL Server's Query Optimizer, Fabiano shines a light on the sixth major Showplan Operator on his list: the Lazy Spool. What does the Lazy Spool do that's so special, how does the Query Optimizer use it, and why is it so Lazy? Fabiano explains all...

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  • Showplan Operator of the Week - Merge Interval

    When Fabiano agreed to undertake the epic task of describing each showplan operator, none of us quite predicted the interesting ways that the series helps to understand how the query optimizer works. With the Merge Interval, Fabiano comes up with some insights about the way that the Query optimizer handles overlapping ranges efficiently. Free trial of SQL Backup™“SQL Backup was able to cut down my backup time significantly AND achieved a 90% compression at the same time!” Joe Cheng. Download a free trial now.

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  • Showplan Operator of the Week - Concatenation

    Fabiano continues in his mission to describe, one week at a time, all the major Showplan Operators used by SQL Server's Query Optimiser to build the Query Plan. This week he gets the Concatenation operator ....Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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  • C# Algorithms for * Operator

    - by Harsha
    I was reading up on Algorithms and came across the Karatsuba multiplication algorithm and a little wiki-ing led to the Schonhage-Strassen and Furer algorithms for multiplication. I was wondering what algorithms are used on the * operator in C#? While multiplying a pair of integers or doubles, does it use a combination of algorithms with some kind of strategy based on the size of the numbers? How could I find out the implementation details for C#?

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  • ShowPlan Operator of the Week - Merge Join

    Did you ever wonder how and why your indexes affect the performances of joins? Once you've read Fabiano Amorim's unforgettable explanation, you'll learn to love the MERGE operator, and plan your indexes so as to allow the Query Optimiser to use it. Free trial of SQL Backup™“SQL Backup was able to cut down my backup time significantly AND achieved a 90% compression at the same time!” Joe Cheng. Download a free trial now.

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  • Call the cast operator of template base class within the derived class

    - by yoni
    I have a template class, called Cell, here the definition: template <class T> class OneCell { ..... } I have a cast operator from Cell to T, here virtual operator const T() const { ..... } Now i have derived class, called DCell, here template <class T> class DCell : public Cell<T> { ..... } I need to override the Cell's cast operator (insert a little if), but after I need to call the Cell's cast operator. In other methods it's should be something like virtual operator const T() const { if (...) { return Cell<T>::operator const T; } else throw ... } but i got a compiler error error: argument of type 'const int (Cell::)()const' does not match 'const int' What can I do? Thank you, and sorry about my poor English.

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  • What are the best practices for implementing the == operator for a class in C#?

    - by remio
    While implementing an == operator, I have the feeling that I am missing some essential points. Hence, I am searching some best practices around that. Here are some related questions I am thinking about: How to cleanly handle the reference comparison? Should it be implemented through a IEquatable<T>-like interface? Or overriding object.Equals? And what about the != operator? (this list might not be exhaustive).

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  • How Does iPhone Visual Voicemail Work From An Operator Perspective?

    - by Jasarien
    I'm hoping there are some Cell Phone Operator gurus here today. Would anyone be able to explain how Operators achieve the Visual Voicemail feature on the iPhone (and I assume other newer smart phones)? If a new cell phone operator that distributed SIM cards wanted to utilise the visual voicemail feature on unlocked iPhone's what services need to be in place to be able to support it? Is there an open spec or is it completely proprietary?

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  • templates and casting operators

    - by Jonathan Swinney
    This code compiles in CodeGear 2009 and Visual Studio 2010 but not gcc. Why? class Foo { public: operator int() const; template <typename T> T get() const { return this->operator T(); } }; Foo::operator int() const { return 5; } The error message is: test.cpp: In member function `T Foo::get() const': test.cpp:6: error: 'const class Foo' has no member named 'operator T'

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  • "AND Operator" in PAM

    - by d_inevitable
    I need to prevent users from authenticating through Kerberos when the encrypted /home/users has not yet been mounted. (This is to avoid corrupting the ecryptfs mountpoint) Currently I have these lines in /etc/pam.d/common-auth: auth required pam_group.so use_first_pass auth [success=2 default=ignore] pam_krb5.so minimum_uid=1000 try_first_pass auth [success=1 default=ignore] pam_unix.so nullok_secure try_first_pass I am planning to use pam_exec.so to execute a script that will exit 1 if the ecyptfs mounts are not ready yet. Doing this: auth required pam_exec.so /etc/security/check_ecryptfs will lock me out for good if ecryptfs for some reason fails. In such case I would like to at least be able to login with a local (non-kerberos) user to fix the issue. Is there some sort of AND-Operator in which I can say that login through kerberos+ldap is only sufficient if both kerberos authentication and the ecryptfs mount has succeeded?

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  • Avoid Postfix Increment Operator

    - by muntoo
    I've read that I should avoid the postfix increment operator because of performance reasons (in certain cases). But doesn't this affect code readability? In my opinion: for(int i = 0; i < 42; i++); /* i will never equal 42! */ Looks better than: for(int i = 0; i < 42; ++i); /* i will never equal 42! */ But this is probably just out of habit. Admittedly, I haven't seen many use ++i. Is the performance that bad to sacrifice readability, in this case? Or am I just blind, and ++i is more readable than i++?

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  • Is Polymorphism and Method Overloading is almost the same thing in C++

    - by Maxood
    In C++, there are 2 types of Polymorphism: Object Polymorphism Function Polymorphism Function polymorphism is exactly the same thing as method or function overloading i.e. We use the same method names with different parameters and return types. Now the question is why do we have this fancy name Polymorphism in OOP? What distinctly distinguishes polymorphism from method overloading? Can someone explain with a scenario. Thanks

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  • Why does virtual assignment behave differently than other virtual functions of the same signature?

    - by David Rodríguez - dribeas
    While playing with implementing a virtual assignment operator I have ended with a funny behavior. It is not a compiler glitch, since g++ 4.1, 4.3 and VS 2005 share the same behavior. Basically, the virtual operator= behaves differently than any other virtual function with respect to the code that is actually being executed. struct Base { virtual Base& f( Base const & ) { std::cout << "Base::f(Base const &)" << std::endl; return *this; } virtual Base& operator=( Base const & ) { std::cout << "Base::operator=(Base const &)" << std::endl; return *this; } }; struct Derived : public Base { virtual Base& f( Base const & ) { std::cout << "Derived::f(Base const &)" << std::endl; return *this; } virtual Base& operator=( Base const & ) { std::cout << "Derived::operator=( Base const & )" << std::endl; return *this; } }; int main() { Derived a, b; a.f( b ); // [0] outputs: Derived::f(Base const &) (expected result) a = b; // [1] outputs: Base::operator=(Base const &) Base & ba = a; Base & bb = b; ba = bb; // [2] outputs: Derived::operator=(Base const &) Derived & da = a; Derived & db = b; da = db; // [3] outputs: Base::operator=(Base const &) ba = da; // [4] outputs: Derived::operator=(Base const &) da = ba; // [5] outputs: Derived::operator=(Base const &) } The effect is that the virtual operator= has a different behavior than any other virtual function with the same signature ([0] compared to [1]), by calling the Base version of the operator when called through real Derived objects ([1]) or Derived references ([3]) while it does perform as a regular virtual function when called through Base references ([2]), or when either the lvalue or rvalue are Base references and the other a Derived reference ([4],[5]). Is there any sensible explanation to this odd behavior?

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  • operator new for array of class without default constructor......

    - by skydoor
    For a class without default constructor, operator new and placement new can be used to declare an array of such class. When I read the code in More Effective C++, I found the code as below(I modified some part)..... My question is, why [] after the operator new is needed? I test it without it, it still works. Can any body explain that? class A { public: int i; A(int i):i(i) {} }; int main() { void *rawMemory = operator new[] (10 * sizeof(A)); // Why [] needed here? A *p = static_cast<A*>(rawMemory); for(int i = 0 ; i < 10 ; i++ ) { new(&p[i])A(i); } for(int i = 0 ; i < 10 ; i++ ) { cout<<p[i].i<<endl; } for(int i = 0 ; i < 10 ; i++ ) { p[i].~A(); } return 0; }

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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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  • Dilemma with two types and operator +

    - by user35443
    I have small problem with operators. I have this code: public class A { public string Name { get; set; } public A() { } public A(string Name) { this.Name = Name; } public static implicit operator B(A a) { return new B(a.Name); } public static A operator+(A a, A b) { return new A(a.Name + " " + b.Name); } } public class B { public string Name { get; set; } public B() { } public B(string Name) { this.Name = Name; } public static implicit operator A(B b) { return new A(b.Name); } public static B operator +(B b, B a) { return new B(b.Name + " " + a.Name); } } Now I want to know, which's conversion operator will be called and which's addition operator will be called in this operation: new A("a") + new B("b"); Will it be operator of A, or of B? (Or both?) Thanks....

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  • How to modify a given class to use const operators

    - by zsero
    I am trying to solve my question regarding using push_back in more than one level. From the comments/answers it is clear that I have to: Create a copy operator which takes a const argument Modify all my operators to const But because this header file is given to me there is an operator what I cannot make into const. It is a simple: float & operator [] (int i) { return _item[i]; } In the given program, this operator is used to get and set data. My problem is that because I need to have this operator in the header file, I cannot turn all the other operators to const, what means I cannot insert a copy operator. How can I make all my operators into const, while preserving the functionality of the already written program? Here is the full declaration of the class: class Vector3f { float _item[3]; public: float & operator [] (int i) { return _item[i]; } Vector3f(float x, float y, float z) { _item[0] = x ; _item[1] = y ; _item[2] = z; }; Vector3f() {}; Vector3f & operator = ( const Vector3f& obj) { _item[0] = obj[0]; _item[1] = obj[1]; _item[2] = obj[2]; return *this; }; Vector3f & operator += ( const Vector3f & obj) { _item[0] += obj[0]; _item[1] += obj[1]; _item[2] += obj[2]; return *this; }; bool operator ==( const Vector3f & obj) { bool x = (_item[0] == obj[0]) && (_item[1] == obj[1]) && (_item[2] == obj[2]); return x; } // my copy operator Vector3f(const Vector3f& obj) { _item[0] += obj[0]; _item[1] += obj[1]; _item[2] += obj[2]; return this; } };

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  • How do I indicate that a class doesn't support certain operators?

    - by romeovs
    I'm writing a class that represents an ordinal scale, but has no logical zero-point (eg time). This scale should permit addition and substraction (operator+, operator+=, ...) but not multiplication. Yet, I always felt it to be a good practice that when one overloads one operator of a certain group (in this case the math operators), one should also overload all the others that belong to that group. In this case that would mean I should need to overload the multiplication and division operators also, because if a user can use A+B he would probable expect to be able the other operators. Is there a method that I can use to throw an error for this at compiler time? The easiest method would be just no to overload the operators operator*, ... yet it would seem appropriate to add a bit more explaination than operator* is not know for class "time". Or is this something that I really should not care about (RTFM user)?

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  • What is the ISO C++ way to directly define a conversion function to reference to array?

    - by ben
    According to the standard, a conversion function has a function-id operator conversion-type-id, which would look like, say, operator char(&)[4] I believe. But I cannot figure out where to put the function parameter list. gcc does not accept either of operator char(&())[4] or operator char(&)[4]() or anything I can think of. Now, gcc seems to accept (&operator char ())[4] but clang does not, and I am inclined to not either, since it does not seem to fit the grammar as I understand it. I do not want to use a typedef because I want to avoid polluting the namespace with it.

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  • DIVIDE vs division operator in #dax

    - by Marco Russo (SQLBI)
    Alberto Ferrari wrote an interesting article about DIVIDE performance in DAX. This new function has been introduced in SQL Server Analysis Services 2012 SP1, so it is available also in Excel 2013 (which still doesn’t have other features/fixes introduced by following Cumulative Updates…). The idea that instead of writing: IF ( Sales[Quantity] <> 0, Sales[Amount] / Sales[Quantity], BLANK () ) you can write: DIVIDE ( Sales[Amount], Sales[Quantity] ) There is a third optional argument in DIVIDE that defines the result in case the denominator (second argument) is zero, and by default its value is BLANK, so I omitted the third argument in my example. Using DIVIDE is very important, especially when you use a measure in MDX (for example in an Excel PivotTable) because it raise the chance that the non empty evaluation for the result is evaluated in bulk mode instead of cell-by-cell. However, from a DAX point of view, you might find it’s better to use the standard division operator removing the IF statement. I suggest you to read Alberto’s article, because you will find that an expression applying a filter using FILTER is faster than using CALCULATE, which is against any rule of thumb you might have read until now! Again, this is not always true, and depends on many conditions – trying to simplify, we might say that for a simple calculation, the query plan generated by FILTER could be more efficient – but, as usual, it depends, and 90% of the times using FILTER instead of CALCULATE produces slower performance. Do not take anything for granted, and always check the query plan when performance are your first issue!

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