Search Results

Search found 2533 results on 102 pages for 'typecast operator'.

Page 7/102 | < Previous Page | 3 4 5 6 7 8 9 10 11 12 13 14  | Next Page >

  • Java operator precedence guidelines

    - by trashgod
    Misunderstanding Java operator precedence is a source of frequently asked questions and subtle errors. I was intrigued to learn that even the Java Language Specification says, "It is recommended that code not rely crucially on this specification." JLS §15.7 Preferring clear to clever, are there any useful guidelines in this area? Here are a number of resources on the topic: JLS Operators JLS Precedence Java Glossary Princeton Sun Tutorial University of West Florida Usenet discussion Additions or corrections welcome.

    Read the article

  • 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

    Read the article

  • 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

    Read the article

  • 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];

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • 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#

    Read the article

  • 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!

    Read the article

  • 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,

    Read the article

  • 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

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

< Previous Page | 3 4 5 6 7 8 9 10 11 12 13 14  | Next Page >