Search Results

Search found 511 results on 21 pages for 'overloading'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 12/21 | < Previous Page | 8 9 10 11 12 13 14 15 16 17 18 19  | Next Page >

  • 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.

    Read the article

  • Why `is_base_of` works with private inheritance?

    - by Alexey Malistov
    Why the following code works? typedef char (&yes)[1]; typedef char (&no)[2]; template <typename B, typename D> struct Host { operator B*() const; operator D*(); }; template <typename B, typename D> struct is_base_of { template <typename T> static yes check(D*, T); static no check(B*, int); static const bool value = sizeof(check(Host<B,D>(), int())) == sizeof(yes); }; //Test sample class B {}; class D : private B {}; //Exspression is true. int test[is_base_of<B,D>::value && !is_base_of<D,B>::value]; Note that B is private base. Note that operator B*() is const. How does this work? Why this works? Why static yes check(D*, T); is better than static yes check(B*, int); ?

    Read the article

  • 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)?

    Read the article

  • c++ specialized overload?

    - by acidzombie24
    -edit- i am trying to close the question. i solved the problem with boost::is_base_and_derived In my class i want to do two things. 1) Copy int, floats and other normal values 2) Copy structs that supply a special copy function (template T copyAs(); } the struct MUST NOT return int's unless i explicitly say ints. I do not want the programmer mistaking the mistake by doing int a = thatClass; -edit- someone mention classes dont return anything, i mean using the operator Type() overload. How do i create my copy operator in such a way i can copy both 1) ints, floats etc and the the struct restricted in the way i mention in 2). i tried doing template <class T2> T operator = (const T2& v); which would cover my ints, floats etc. But how would it differentiate from structs? so i wrote T operator = (const SomeGenericBase& v); The idea was the GenericBase would be unsed instead then i can do v.Whatever. But that backfires bc the functions i want wouldnt exist, unless i use virtual, but virtual templates dont exist. Also i would hate to use virtual I think the solution is to get rid of ints and have it convert to something that can do .as(). So i wrote something up but now i have the same problem, how does that differentiate ints and structs that have the .as() function template?

    Read the article

  • 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.

    Read the article

  • [Netbeans 6.9] Java MethodOverloading error with double values

    - by Nimitips
    Here is a part of my code I'm having trouble with: ===Class Overload=== public class Overload { public void testOverLoadeds() { System.out.printf("Square of integer 7 is %d\n",square(7)); System.out.printf("Square of double 7.5 is %d\n",square(7.5)); }//..end testOverloadeds public int square(int intValue) { System.out. printf("\nCalled square with int argument: %d\n",intValue); return intValue * intValue; }//..end square int public double square(double doubleValue) { System.out.printf("\nCalled square with double argument: %d\n", doubleValue); return doubleValue * doubleValue; }//..end square double }//..end class overload ===Main=== public static void main(String[] args) { Overload methodOverload = new Overload(); methodOverload.testOverLoadeds(); } It compiles with no error, however when I try to run it the output is: Called square with int argument: 7 Square of integer 7 is 49 Exception in thread "main" java.util.IllegalFormatConversionException: d != java.lang.Double at java.util.Formatter$FormatSpecifier.failConversion(Formatter.java:3999) at java.util.Formatter$FormatSpecifier.printInteger(Formatter.java:2709) at java.util.Formatter$FormatSpecifier.print(Formatter.java:2661) at java.util.Formatter.format(Formatter.java:2433) at java.io.PrintStream.format(PrintStream.java:920) at java.io.PrintStream.printf(PrintStream.java:821) at methodoverload.Overload.square(Overload.java:19) at methodoverload.Overload.testOverLoadeds(Overload.java:8) at methodoverload.Main.main(Main.java:9) Called square with double argument:Java Result: 1 What am I doing wrong? I'm on Ubuntu 10.10, Netbeans 6.9. Thanks.

    Read the article

  • How do you override operator == when using interfaces instead of actual types?

    - by RickL
    I have some code like this: How should I implement the operator == so that it will be called when the variables are of interface IMyClass? public class MyClass : IMyClass { public static bool operator ==(MyClass a, MyClass b) { if (ReferenceEquals(a, b)) return true; if ((Object)a == null || (Object)b == null) return false; return false; } public static bool operator !=(MyClass a, MyClass b) { return !(a == b); } } class Program { static void Main(string[] args) { IMyClass m1 = new MyClass(); IMyClass m2 = new MyClass(); MyClass m3 = new MyClass(); MyClass m4 = new MyClass(); Console.WriteLine(m1 == m2); // does not go into custom == function. why not? Console.WriteLine(m3 == m4); // DOES go into custom == function } }

    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

  • perl - universal operator overload

    - by Todd Freed
    I have an idea for perl, and I'm trying to figure out the best way to implement it. The idea is to have new versions of every operator which consider the undefined value as the identity of that operation. For example: $a = undef + 5; # undef treated as 0, so $a = 5 $a = undef . "foo"; # undef treated as '', so $a = foo $a = undef && 1; # undef treated as false, $a = true and so forth. ideally, this would be in the language as a pragma, or something. use operators::awesome; However, I would be satisfied if I could implement this special logic myself, and then invoke it where needed: use My::Operators; The problem is that if I say "use overload" inside My::Operators only affects objects blessed into My::Operators. So the question is: is there a way (with "use overoad" or otherwise) to do a "universal operator overload" - which would be called for all operations, not just operations on blessed scalars. If not - who thinks this would be a great idea !? It would save me a TON of this kind of code if($object && $object{value} && $object{value} == 15) replace with if($object{value} == 15) ## the special "is-equal-to" operator

    Read the article

  • What is operator<< <> in C++?

    - by Austin Hyde
    I have seen this in a few places, and to confirm I wasn't crazy, I looked for other examples. Apparently this can come in other flavors as well, eg operator+ <>. However, nothing I have seen anywhere mentions what it is, so I thought I'd ask. It's not the easiest thing to google operator<< <>( :-)

    Read the article

  • When I overload the assignment operator for my simple class array, I get the wrong answer I espect

    - by user299648
    //output is "01234 00000" but the output should be or what I want it to be is // "01234 01234" because of the assignment overloaded operator #include <iostream> using namespace std; class IntArray { public: IntArray() : size(10), used(0) { a= new int[10]; } IntArray(int s) : size(s), used(0) { a= new int[s]; } int& operator[]( int index ); IntArray& operator =( const IntArray& rightside ); ~IntArray() { delete [] a; } private: int *a; int size; int used;//for array position }; int main() { IntArray copy; if( 2>1) { IntArray arr(5); for( int k=0; k<5; k++) arr[k]=k; copy = arr; for( int j=0; j<5; j++) cout<<arr[j]; } cout<<" "; for( int j=0; j<5; j++) cout<<copy[j]; return 0; } int& IntArray::operator[]( int index ) { if( index >= size ) cout<<"ilegal index in IntArray"<<endl; return a[index]; } IntArray& IntArray::operator =( const IntArray& rightside ) { if( size != rightside.size )//also checks if on both side same object { delete [] a; a= new int[rightside.size]; } size=rightside.size; used=rightside.used; for( int i = 0; i < used; i++ ) a[i]=rightside.a[i]; return *this; }

    Read the article

  • Why friend function is preferred to member function for operator<<

    - by skydoor
    When you are going to print an object, a friend operator<< is used. Can we use member function for operator<< ? class A { public: void operator<<(ostream& i) { i<<"Member function";} friend ostream& operator<<(ostream& i, A& a) { i<<"operator<<"; return i;} }; int main () { A a; A b; A c; cout<<a<<b<<c<<endl; a<<cout; return 0; } One point is that friend function enable us to use it like this cout<<a<<b<<c What other reasons?

    Read the article

  • Why is T() = T() allowed?

    - by Rimo
    I believe the expression T() creates an rvalue (by the Standard). However, the following code compiles (at least on gcc4.0): class T {}; int main() { T() = T(); } I know technically this is possible because member functions can be invoked on temporaries and the above is just invoking the operator= on the rvalue temporary created from the first T(). But conceptually this is like assigning a new value to an rvalue. Is there a good reason why this is allowed? Edit: The reason I find this odd is it's strictly forbidden on built-in types yet allowed on user-defined types. For example, int(2) = int(3) won't compile because that is an "invalid lvalue in assignment". So I guess the real question is, was this somewhat inconsistent behavior built into the language for a reason? Or is it there for some historical reason? (E.g it would be conceptually more sound to allow only const member functions to be invoked on rvalue expressions, but that cannot be done because that might break some existing code.)

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • Comparing objects and inheritance

    - by ereOn
    Hi, In my program I have the following class hierarchy: class Base // Base is an abstract class { }; class A : public Base { }; class B : public Base { }; I would like to do the following: foo(const Base& one, const Base& two) { if (one == two) { // Do something } else { // Do something else } } I have issues regarding the operator==() here. Of course comparing an instance A and an instance of B makes no sense but comparing two instances of Base should be possible. (You can't compare a Dog and a Cat however you can compare two Animals) I would like the following results: A == B = false A == A = true or false, depending on the effective value of the two instances B == B = true or false, depending on the effective value of the two instances My question is: is this a good design/idea ? Is this even possible ? What functions should I write/overload ? My apologies if the question is obviously stupid or easy, I have some serious fever right now and my thinking abilities are somewhat limited :/ Thank you.

    Read the article

  • C++: Overload != When == Overloaded

    - by Mark W
    Say I have a class where I overloaded the operator == as such: Class A { ... public: bool operator== (const A &rhs) const; ... }; ... bool A::operator== (const A &rhs) const { .. return isEqual; } I already have the operator == return the proper Boolean value. Now I want to extend this to the simple opposite (!=). I would like to call the overloaded == operator and return the opposite, i.e. something of the nature bool A::operator!= (const A &rhs) const { return !( this == A ); } Is this possible? I know this will not work, but it exemplifies what I would like to have. I would like to keep only one parameter for the call: rhs. Any help would be appreciated, because I could not come up with an answer after several search attempts.

    Read the article

  • Why does the Scala compiler disallow overloaded methods with default arguments?

    - by soc
    While there might be valid cases where such method overloadings could become ambiguous, why does the compiler disallow code which is neither ambiguous at compile time nor at run time? Example: // This fails: def foo(a: String)(b: Int = 42) = a + b def foo(a: Int) (b: Int = 42) = a + b // This fails, too. Even if there is no position in the argument list, // where the types are the same. def foo(a: Int) (b: Int = 42) = a + b def foo(a: String)(b: String = "Foo") = a + b // This is OK: def foo(a: String)(b: Int) = a + b def foo(a: Int) (b: Int = 42) = a + b // Even this is OK. def foo(a: Int)(b: Int) = a + b def foo(a: Int)(b: String = "Foo") = a + b val bar = foo(42)_ // This complains obviously ... Are there any reasons why these restrictions can't be loosened a bit? Especially when converting heavily overloaded Java code to Scala default arguments are a very important and it isn't nice to find out after replacing plenty of Java methods by one Scala methods that the spec/compiler imposes arbitrary restrictions.

    Read the article

  • Why can operator-> be overloaded manually?

    - by FredOverflow
    Wouldn't it make sense if p->m was just syntactic sugar for (*p).m? Essentially, every operator-> that I have ever written could have been implemented as follows: Foo::Foo* operator->() { return &**this; } Is there any case where I would want p->m to mean something else than (*p).m?

    Read the article

  • copy C'tor with operator= | C++

    - by user2266935
    I've got this code here: class DerivedClass : public BaseClass { SomeClass* a1; Someclass* a2; public: //constructors go here ~DerivedClass() { delete a1; delete a2;} // other functions go here .... }; My first question is as follows: Can I write an "operator=" to "DerivedClass" ? (if your answer is yes, could you show me how?) My second question is: If the answer to the above is yes, could you show me how to make an "copy c'tor" using the "operator=" that you wrote beforehand (if that is even possible)? Your help would be much appreciated !

    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

  • signature output operator overload

    - by coubeatczech
    hi, do you know, how to write signature of a function or method for operator<< for template class in C++? I want something like: template <class A class MyClass{ public: friend ostream & operator<<(ostream & os, MyClass<A mc); } ostream & operator<<(ostream & os, MyClass<A mc){ // some code return os; } But this just won't compile. Do anyone know, how to write it correctly?

    Read the article

  • Prolog: declaring an operator

    - by B K
    I have defined ! (factorial) function and registered it as arithmetic function and an operator, so that I can execute: A is 6!. Now I'd like to define !! (factorial of odd numbers), but the same way - writing clauses, registering arithmetic_function and operator, calling A is 7!! - results in SyntaxError: Operator expected How should I, if possible, register !! operator ? Yes, I realize, ! is normally the cut.

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 8 9 10 11 12 13 14 15 16 17 18 19  | Next Page >