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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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  • Operator precedence in scala

    - by Jeriho
    I like scala's propose of operator precedence but in some rare case unmodified rules may be inconvenient because you have restrictions in naming your methods. Is there in scala ways to define another rules for a class/file/etc? If not would it be resolved in future?

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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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  • Union,Except and Intersect operator in Linq

    - by Jalpesh P. Vadgama
    While developing a windows service using Linq-To-SQL i was in need of something that will intersect the two list and return a list with the result. After searching on net i have found three great use full operators in Linq Union,Except and Intersect. Here are explanation of each operator. Union Operator: Union operator will combine elements of both entity and return result as third new entities. Except Operator: Except operator will remove elements of first entities which elements are there in second entities and will return as third new entities. Intersect Operator: As name suggest it will return common elements of both entities and return result as new entities. Let’s take a simple console application as  a example where i have used two string array and applied the three operator one by one and print the result using Console.Writeline. Here is the code for that. C#, using GeSHi 1.0.8.6 using System; using System.Collections.Generic; using System.Linq; using System.Text;     namespace ConsoleApplication1 {     class Program     {         static void Main(string[] args)         {             string[] a = { "a", "b", "c", "d" };             string[] b = { "d","e","f","g"};               var UnResult = a.Union(b);             Console.WriteLine("Union Result");               foreach (string s in UnResult)             {                 Console.WriteLine(s);                          }               var ExResult = a.Except(b);             Console.WriteLine("Except Result");             foreach (string s in ExResult)             {                 Console.WriteLine(s);             }               var InResult = a.Intersect(b);             Console.WriteLine("Intersect Result");             foreach (string s in InResult)             {                 Console.WriteLine(s);             }             Console.ReadLine();                        }          } }   Parsed in 0.022 seconds at 45.54 KB/s Here is the output of console application as Expected. Hope this will help you.. Technorati Tags: Linq,Except,InterSect,Union,C#

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  • Using LIKE operator in LINQ to Entity

    - by Draconic
    Hi, everybody! Currently in our project we are using Entity Framework and LINQ. We want to create a search feature where the Client fills different filters but he isn't forced to. To do this "dynamic" query in LINQ, we thought about using the Like operator, searching either for the field, or "%" to get everything if the user didn't fill that field. The joke's on us when we discovered it didn't support Like. After some searching, we read several answers where it's sugested to use StartsWith, but it's useless for us. Is the only solution using something like: ObjectQuery<Contact> contacts = db.Contacts; if (pattern != "") { contacts = contacts.Where(“it.Name LIKE @pattern”); contacts.Parameters.Add(new ObjectParameter(“pattern”, pattern); } However, we'd like to stick with linq only. Happy coding!

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  • What division operator symbol would you pick?

    - by Mackenzie
    I am currently designing and implementing a small programming language as an extra-credit project in a class I'm taking. My problem is that the language has three numeric types: Long, Double, and Fraction. Fractions can be written in the language as proper or improper fractions (e.g. "2 1/3" or "1/2"). This fact leads to problems such as "2/3.5" (Long/Double) and "2/3"(Long/Long) not being handled correctly by the lexer.The best solution that I see is to change the division operator. So far, I think "\" is the best solution since "//" starts comments. Would you pick "\", if you were designing the language? Would you pick something else? If so, what? Note: changing the way fractions are written is not possible. Thanks in advance for your help,

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  • F# operator over-loading question

    - by jyoung
    The following code fails in 'Evaluate' with: "This expression was expected to have type Complex but here has type double list" Am I breaking some rule on operator over-loading on '(+)'? Things are OK if I change '(+)' to 'Add'. open Microsoft.FSharp.Math /// real power series [kn; ...; k0] => kn*S^n + ... + k0*S^0 type Powers = double List let (+) (ls:Powers) (rs:Powers) = let rec AddReversed (ls:Powers) (rs:Powers) = match ( ls, rs ) with | ( l::ltail, r::rtail ) -> ( l + r ) :: AddReversed ltail rtail | ([], _) -> rs | (_, []) -> ls ( AddReversed ( ls |> List.rev ) ( rs |> List.rev) ) |> List.rev let Evaluate (ks:Powers) ( value:Complex ) = ks |> List.fold (fun (acc:Complex) (k:double)-> acc * value + Complex.Create(k, 0.0) ) Complex.Zero

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  • STL: how to overload operator= for <vector> ?

    - by MBes
    There's simple example: #include <vector> int main() { vector<int> veci; vector<double> vecd; for(int i = 0;i<10;++i){ veci.push_back(i); vecd.push_back(i); } vecd = veci; // <- THE PROBLEM } The thing I need to know is how to overload operator = so that I could make assignment like this: vector<double> = vector<int>; I've just tried a lot of ways, but always compiler has been returning errors... Is there any option to make this code work without changing it? I can write some additional lines, but can't edit or delete the existing ones. Ty.

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  • C# string equality operator returns false, but I'm pretty sure it should be true... What?!

    - by Daniel Schaffer
    I'm trying to write a unit test for a piece of code that generates a large amount of text. I've run into an issue where the "expected" and "actual" strings appear to be equal, but Assert.AreEqual throws, and both the equality operator and Equals() return false. The result of GetHashCode() is different for both values as well. However, putting both strings into text files and comparing with DiffMerge tells me they're the same. Additionally, using Encoding.ASCII.GetBytes() on both values and then using SequenceEquals to compare the resulting byte arrays returns true. The values are 34KB each, so I'll hold off putting them here for now. Any ideas? I'm completely stumped.

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  • Specification Pattern and Boolean Operator Precedence

    - by Anders Nielsen
    In our project, we have implemented the Specification Pattern with boolean operators (see DDD p 274), like so: public abstract class Rule { public Rule and(Rule rule) { return new AndRule(this, rule); } public Rule or(Rule rule) { return new OrRule(this, rule); } public Rule not() { return new NotRule(this); } public abstract boolean isSatisfied(T obj); } class AndRule extends Rule { private Rule one; private Rule two; AndRule(Rule one, Rule two) { this.one = one; this.two = two; } public boolean isSatisfied(T obj) { return one.isSatisfied(obj) && two.isSatisfied(obj); } } class OrRule extends Rule { private Rule one; private Rule two; OrRule(Rule one, Rule two) { this.one = one; this.two = two; } public boolean isSatisfied(T obj) { return one.isSatisfied(obj) || two.isSatisfied(obj); } } class NotRule extends Rule { private Rule rule; NotRule(Rule obj) { this.rule = obj; } public boolean isSatisfied(T obj) { return !rule.isSatisfied(obj); } } Which permits a nice expressiveness of the rules using method-chaining, but it doesn't support the standard operator precedence rules of which can lead to subtle errors. The following rules are not equivalent: Rule<Car> isNiceCar = isRed.and(isConvertible).or(isFerrari); Rule<Car> isNiceCar2 = isFerrari.or(isRed).and(isConvertible); The rule isNiceCar2 is not satisfied if the car is not a convertible, which can be confusing since if they were booleans isRed && isConvertible || isFerrari would be equivalent to isFerrari || isRed && isConvertible I realize that they would be equivalent if we rewrote isNiceCar2 to be isFerrari.or(isRed.and(isConvertible)), but both are syntactically correct. The best solution we can come up with, is to outlaw the method-chaining, and use constructors instead: OR(isFerrari, AND(isConvertible, isRed)) Does anyone have a better suggestion?

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  • Java operator overloading

    - by nimcap
    Not using operators makes my code obscure. (aNumber / aNother) * count is better than aNumber.divideBy(aNother).times(count) After 6 months of not writing a single comment I had to write a comment to the simple operation above. Usually I refactor until I don't need comment. And this made me realize that it is easier to read and perceive math symbols and numbers than their written forms. For example TWENTY_THOUSAND_THIRTEEN.plus(FORTY_TWO.times(TWO_HUNDERED_SIXTY_ONE)) is more obscure than 20013 + 42*261 So do you know a way to get rid of obscurity while not using operator overloading in Java? Update: I did not think my exaggeration on comments would cause such trouble to me. I am admitting that I needed to write comment a couple of times in 6 months. But not more than 10 lines in total. Sorry for that. Update 2: Another example: budget.plus(bonusCoefficient.times(points)) is more obscure than budget + bonusCoefficient * points I have to stop and think on the first one, at first sight it looks like clutter of words, on the other hand, I get the meaning at first look for the second one, it is very clear and neat. I know this cannot be achieved in Java but I wanted to hear some ideas about my alternatives.

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  • Constructor or Assignment Operator

    - by ju
    Can you help me is there definition in C++ standard that describes which one will be called constructor or assignment operator in this case: #include <iostream> using namespace std; class CTest { public: CTest() : m_nTest(0) { cout << "Default constructor" << endl; } CTest(int a) : m_nTest(a) { cout << "Int constructor" << endl; } CTest(const CTest& obj) { m_nTest = obj.m_nTest; cout << "Copy constructor" << endl; } CTest& operatorint rhs) { m_nTest = rhs; cout << "Assignment" << endl; return *this; } protected: int m_nTest; }; int _tmain(int argc, _TCHAR* argv[]) { CTest b = 5; return 0; } Or is it just a matter of compiler optimization?

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  • Conditional operator in Mako using Pylons

    - by Antoine Leclair
    In PHP, I often use the conditional operator to add an attribute to an html element if it applies to the element in question. For example: <select name="blah"> <option value="1"<?= $blah == 1 ? ' selected="selected"' : '' ?>> One </option> <option value="2"<?= $blah == 2 ? ' selected="selected"' : '' ?>> Two </option> </select> I'm starting a project with Pylons using Mako for the templating. How can I achieve something similar? Right now, I see two possibilities that are not ideal. Solution 1: <select name="blah"> % if blah == 1: <option value="1" selected="selected">One</option> % else: <option value="1">One</option> % endif % if blah == 2: <option value="2" selected="selected">Two</option> % else: <option value="2">Two</option> % endif </select> Solution 2: <select name="blah"> <option value="1" % if blah == 1: selected="selected" % endif >One</option> <option value="2" % if blah == 2: selected="selected" % endif >Two</option> </select> In this particular case, the value is equal to the variable tested (value="1" = blah == 1), but I use the same pattern in other situations, like <?= isset($variable) ? ' value="$variable" : '' ?>. I am looking for a clean way to achieve this using Mako.

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  • Type result with Ternary operator in C#

    - by Vaccano
    I am trying to use the ternary operator, but I am getting hung up on the type it thinks the result should be. Below is an example that I have contrived to show the issue I am having: class Program { public static void OutputDateTime(DateTime? datetime) { Console.WriteLine(datetime); } public static bool IsDateTimeHappy(DateTime datetime) { if (DateTime.Compare(datetime, DateTime.Parse("1/1")) == 0) return true; return false; } static void Main(string[] args) { DateTime myDateTime = DateTime.Now; OutputDateTime(IsDateTimeHappy(myDateTime) ? null : myDateTime); Console.ReadLine(); ^ } | } | // This line has the compile issue ---------------+ On the line indicated above, I get the following compile error: Type of conditional expression cannot be determined because there is no implicit conversion between '< null ' and 'System.DateTime' I am confused because the parameter is a nullable type (DateTime?). Why does it need to convert at all? If it is null then use that, if it is a date time then use that. I was under the impression that: condition ? first_expression : second_expression; was the same as: if (condition) first_expression; else second_expression; Clearly this is not the case. What is the reasoning behind this? (NOTE: I know that if I make "myDateTime" a nullable DateTime then it will work. But why does it need it? As I stated earlier this is a contrived example. In my real example "myDateTime" is a data mapped value that cannot be made nullable.)

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  • Efficiency of manually written loops vs operator overloads (C++)

    - by Sagekilla
    Hi all, in the program I'm working on I have 3-element arrays, which I use as mathematical vectors for all intents and purposes. Through the course of writing my code, I was tempted to just roll my own Vector class with simple +, -, *, /, etc overloads so I can simplify statements like: for (int i = 0; i < 3; i++) r[i] = r1[i] - r2[i]; // becomes: r = r1 - r2; Which should be more or less identical in generated code. But when it comes to more complicated things, could this really impact my performance heavily? One example that I have in my code is this: Manually written version: for (int j = 0; j < 3; j++) { p.vel[j] = p.oldVel[j] + (p.oldAcc[j] + p.acc[j]) * dt2 + (p.oldJerk[j] - p.jerk[j]) * dt12; p.pos[j] = p.oldPos[j] + (p.oldVel[j] + p.vel[j]) * dt2 + (p.oldAcc[j] - p.acc[j]) * dt12; } Using a Vector class with operator overloads: p.vel = p.oldVel + (p.oldAcc + p.acc) * dt2 + (p.oldJerk - p.jerk) * dt12; p.pos = p.oldPos + (p.oldVel + p.vel) * dt2 + (p.oldAcc - p.acc) * dt12; I am compiling my code for maximum possible speed, as it's extremely important that this code runs quickly and calculates accurately. So will me relying on my Vector's for these sorts of things really affect me? For those curious, this is part of some numerical integration code which is not trivial to run in my program. Any insight would be appreciated, as would any idioms or tricks I'm unaware of.

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  • Pair equal operator overloading for inserting into set

    - by Petwoip
    I am trying to add a pair<int,int> to a set. If a pair shares the same two values as another in the set, it should not be inserted. Here's my non-working code: typedef std::pair<int, int> PairInt; template<> bool std::operator==(const PairInt& l, const PairInt& r) { return (l.first == r.first && l.second == r.second) || (l.first == r.second && l.second == r.first); } int main() { std::set<PairInt> intSet; intSet.insert(PairInt(1,3)); intSet.insert(PairInt(1,4)); intSet.insert(PairInt(1,4)); intSet.insert(PairInt(4,1)); } At the moment, the (4,1) pair gets added even though there is already a (1,4) pair. The final contents of the set are: (1 3) (1 4) (4 1) and I want it to be (1 3) (1 4) I've tried putting breakpoints in the overloaded method, but they never get reached. What have I done wrong?

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  • Equality and Assigment Operators

    - by Jeremy Smith
    I have a assembly compiled in VB.NET that contains two operators: Public Shared Operator =(quarterA As CalendarQuarter, quarterB As CalendarQuarter) As Boolean Return quarterA.StartDate = quarterB.StartDate AndAlso quarterA.EndDate = quarterB.EndDate AndAlso quarterA.Quarter = quarterB.Quarter End Operator Public Shared Operator <>(quarterA As CalendarQuarter, quarterB As CalendarQuarter) As Boolean Return Not (quarterA = quarterB) End Operator However, when using the assembly in C# to perform equality checks if (qtr != null) I receive the error: Cannot implicity convert type 'object' to 'bool' My original intent with the = operator was only for assignment purposes in VB, so I may be way off base (I don't use custom operators too often). What do I need to do to make the operator behave with both equality and assignment operations?

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  • C++ OOP - Can you 'overload a cast' <- hard to explain in 1 sentence

    - by Brandon Miller
    Well, the WinAPI has a POINT struct, but I am trying to make an alternative class to this so you can set the values of x and y from a constructor. /** * X-Y coordinates */ class Point { public: int X, Y; Point(void) : X(0), Y(0) {} Point(int x, int y) : X(x), Y(y) {} Point(const POINT& pt) : X(pt.x), Y(pt.y) {} Point& operator= (const POINT& other) { X = other.x; Y = other.y; } }; // I have an assignment operator and copy constructor. Point myPtA(3,7); Point myPtB(8,5); POINT pt; pt.x = 9; pt.y = 2; // I can assign a 'POINT' to a 'Point' myPtA = pt; // But I also want to be able to assign a 'Point' to a 'POINT' pt = myPtB; Is it possible to overload operator= in a way so that I can assign a Point to a POINT? Or maybe some other method to achieve this? Thanks in advance.

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  • ASX: Just Another Stock Market Operator

    - by Theresa Hickman
    I try to stay informed with what's happening in global financial markets since we all know they are all interconnected. Last week, on Mar. 11 2010, Australia's Senate passed a law that reduced Australia's stock market's role to just a stock market operator. Before this, ASX (Australian Stock Exchange) acted as both its own regulator and operator (supervising trade actvities and handling the trades) of Australia's stock market. Many viewed this as a conflict of interest. So now, the Australian Securities & Investments Commision (ASIC) will act as regulator and ASX will simply be a stock market operator to ensure the continued integrity of financial markets. I believe what this is doing is laying the groundwork to have more than one stock exchange in Australia. I woudn't be surpised if Nasdaq makes a play. As you may or may not know, Nasdaq had been trying for years to take over control of the London Stock Exchange (LSE), which LSE had rejected because it thinks it is worth more than what Nasdaq is willing to pay. Nasdaq or even NYSE may want a piece of Asia/Pacific because nowadays most of the IPOs are coming from foreign companies outside the US. I didn't know this, but apparently many Asia/Pacific stock exchanges have a monopoly where they act as both regulator and operator. I'll be curious to see what happens after the ASIC meet and decide how to regulate Australia's stock exchange to see how many suitors come running towards Australia's financial market.

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  • Nullable types and ?? operator C# [en-US]

    - by ruimachado
    Nullable types vs Non-nullable types   While developing our C# projects its frequent the null comparison operation to avoid null exceptions. This simple operation is mainly coded using the "var x = null" code example inside an if clause. However not all types of variables are nullable, which means that setting a variable to null is not allowed in every cases, it depends on what kind of type are you defining. But what if there was an extension to your non-nullable type that would convert your variable types to nullable? This extension really exists. As I said before in C# you have nullable types which represent all the values of an underlying type, and an additional null value and can be declared easily using "T?", where T is the type of the variable and for example the normal int type cannot be null, so its a non-nullable type, however if you define a "int?" your variable can be null, what you do is convert a non-nullable type to a nullable type. Example: int x=null;     Not allowed     int? x=null;   Allowed     While using nullable types you can check if a variable is null the same way you do it with nullable types:     But what about setting a default value when a certain variable is null?   In this cases the c# .net framework let you set a default value when you try to assign a nullable type to a non-nullable type, using the ?? operator. If you don't use this operator you can still catch the InvalidOperationException which is throw in this cases. For example  without the ?? operator :     Using the ?? operator your code becomes cleaner and more easy to read and you get a bonus, you can set a default value for multiple variables using the ?? in a chain set.     That’s it,   Thanks, Rui Machado rpmachado.wordpress.com

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  • C# ?? null coalescing operator

    - by anirudha
    the null coalescing operator is used for set the value when object is null. if object have some value that nothing change and still have their default value they have.  string str = "i am string";            string message = str ?? "it is null";   the message have same value as str variable because str not null. if str is null that message have value “it is null”; as declared in statement. coalescing operator does not work on nullable operator such as int?

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