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  • Assigning a vector of one type to a vector of another type

    - by deworde
    Hi, I have an "Event" class. Due to the way dates are handled, we need to wrap this class in a "UIEvent" class, which holds the Event, and the date of the Event in another format. What is the best way of allowing conversion from Event to UIEvent and back? I thought overloading the assignment or copy constructor of UIEvent to accept Events (and vice versa)might be best.

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  • Is calling of overload operator-> resolved at compile time?

    - by Brent
    when I tried to compile the code: (note: func and func2 is not typo) struct S { void func2() {} }; class O { public: inline S* operator->() const; private: S* ses; }; inline S* O::operator->() const { return ses; } int main() { O object; object->func(); return 0; } there is a compile error reported: D:\code>g++ operatorp.cpp -S -o operatorp.exe operatorp.cpp: In function `int main()': operatorp.cpp:27: error: 'struct S' has no member named 'func' it seems that invoke the overloaded function of "operator-" is done during compile time? I'd added "-S" option for compile only.

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  • Is it possible to supply template parameters when calling operator()?

    - by Paul
    I'd like to use a template operator() but am not sure if it's possible. Here is a simple test case that won't compile. Is there something wrong with my syntax, or is this simply not possible? struct A { template<typename T> void f() { } template<typename T> void operator()() { } }; int main() { A a; a.f<int>(); // This compiles. a.operator()<int>(); // This compiles. a<int>(); // This won't compile. return 0; }

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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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  • "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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  • 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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  • Where to add an overloaded operator for the tr1::array?

    - by phlipsy
    Since I need to add an operator& for the std::tr1::array<bool, N> I wrote the following lines template<std::size_t N> std::tr1::array<bool, N> operator& (const std::tr1::array<bool, N>& a, const std::tr1::array<bool, N>& b) { std::tr1::array<bool, N> result; std::transform(a.begin(), a.end(), b.begin(), result.begin(), std::logical_and<bool>()); return result; } Now I don't know in which namespace I've to put this function. I considered the std namespace as a restricted area. Only total specialization and overloaded function templates are allowed to be added by the user. Putting it into the global namespace isn't "allowed" either in order to prevent pollution of the global namespace and clashes with other declarations. And finally putting this function into the namespace of the project doesn't work since the compiler won't find it there. What had I best do? I don't want to write a new array class putted into the project namespace. Because in this case the compiler would find the right namespace via argument dependent name lookup. Or is this the only possible way because writing a new operator for existing classes means extending their interfaces and this isn't allowed either for standard classes?

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  • Operator overloading in generic struct: can I create overloads for specific kinds(?) of generic?

    - by Carson Myers
    I'm defining physical units in C#, using generic structs, and it was going okay until I got the error: One of the parameters of a binary operator must be the containing type when trying to overload the mathematical operators so that they convert between different units. So, I have something like this: public interface ScalarUnit { } public class Duration : ScalarUnit { } public struct Scalar<T> where T : ScalarUnit { public readonly double Value; public Scalar(double Value) { this.Value = Value; } public static implicit operator double(Scalar<T> Value) { return Value.Value; } } public interface VectorUnit { } public class Displacement : VectorUnit { } public class Velocity : VectorUnit { } public struct Vector<T> where T : VectorUnit { #... public static Vector<Velocity> operator /(Vector<Displacement> v1, Scalar<Duration> v2) { return new Vector<Velocity>(v1.Magnitude / v2, v1.Direction); } } There aren't any errors for the + and - operators, where I'm just working on a Vector<T>, but when I substitute a unit for T, suddenly it doesn't like it. Is there a way to make this work? I figured it would work, since Displacement implements the VectorUnit interface, and I have where T : VectorUnit in the struct header. Am I at least on the right track here? I'm new to C# so I have difficulty understanding what's going on sometimes.

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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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