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  • 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<< <>( :-)

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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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  • 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.)

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  • 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.

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  • 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?

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  • 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.

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  • 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.

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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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  • 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?

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  • Make conversion to a native type explicit in C++

    - by Tal Pressman
    I'm trying to write a class that implements 64-bit ints for a compiler that doesn't support long long, to be used in existing code. Basically, I should be able to have a typedef somewhere that selects whether I want to use long long or my class, and everything else should compile and work. So, I obviously need conversion constructors from int, long, etc., and the respective conversion operators (casts) to those types. This seems to cause errors with arithmetic operators. With native types, the compiler "knows" that when operator*(int, char) is called, it should promote the char to int and call operator*(int, int) (rather than casting the int to char, for example). In my case it gets confused between the various built-in operators and the ones I created. It seems to me like if I could flag the conversion operators as explicit somehow, that it would solve the issue, but as far as I can tell the explicit keyword is only for constructors (and I can't make constructors for built-in types). So is there any way of marking the casts as explicit? Or am I barking up the wrong tree here and there's another way of solving this? Or maybe I'm just doing something else wrong...

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  • MS SQL tuning tools for finding overload

    - by SkyFox
    I use MS SQL server as a DBMS for my very big corporate DB (with different financial data). And some times my system go down. I don't understand why. What programs/tools I can use for finding process/program/thread, that overload my SQL-server? Thanks for all answers!

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  • Overload and hide methods in Java

    - by Marco
    Hi, i have an abstract class BaseClass with a public insert() method: public abstract class BaseClass { public void insert(Object object) { // Do something } } which is extended by many other classes. For some of those classes, however, the insert() method must have additional parameters, so that they instead of overriding it I overload the method of the base class with the parameters required, for example: public class SampleClass extends BaseClass { public void insert(Object object, Long param){ // Do Something } } Now, if i instantiate the SampleClass class, i have two insert() methods: SampleClass sampleClass = new SampleClass(); sampleClass.insert(Object object); sampleClass.insert(Object object, Long param); what i'd like to do is to hide the insert() method defined in the base class, so that just the overload would be visible: SampleClass sampleClass = new SampleClass(); sampleClass.insert(Object object, Long param); Could this be done in OOP?

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  • 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.

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  • Backward compatibility in dlls

    - by michaelleuzinger
    Hi I do have three dlls. a.dll - released many years ago b.dll - released not so many years c.dll - released shortly Each one contains the same function - unfortunatelly with different parameters. so I do have the following Methods aMethod(param1) aMethod(param1, param2) aMethod(param1, param2, param3) My Task is to make a new dll (or new dlls) wich is backward compatible. But as far as I've learned from Google there is no possibility to overload methods in a dll. Does any one have a tip how I can solve this problem elegantly? -- Michael

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  • Overload with different return type in java?

    - by nunos
    So, I am just starting Java and, even though I have looked in some question about it here at stackoverflow.com and elsewhere, haven't been able to find a straightforward answer to why isn't possible to overload a function just by changing the return type. Why is it so? Will that provably change in a future version of Java? By the way, just for reference, is this possible in C++? Thanks.

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  • Is it possible to implement events in C++?

    - by acidzombie24
    I wanted to implement a C# event in C++ just to see if i could do it. I got stuck, i know the bottom is wrong but what i realize my biggest problem is... How do i overload the () operator to be whatever is in T in this case int func(float)? I cant? can i? Can i implement a good alternative? #include <deque> using namespace std; typedef int(*MyFunc)(float); template<class T> class MyEvent { deque<T> ls; public: MyEvent& operator +=(T t) { ls.push_back(t); return *this; } }; static int test(float f){return (int)f; } int main(){ MyEvent<MyFunc> e; e += test; }

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  • template; operator (int)

    - by Oops
    Hi, regarding my Point struct already mentioned here: http://stackoverflow.com/questions/2794369/template-class-ctor-against-function-new-c-standard is there a chance to replace the function toint() with a cast-operator (int)? namespace point { template < unsigned int dims, typename T > struct Point { T X[ dims ]; //umm??? template < typename U > Point< dims, U > operator U() const { Point< dims, U > ret; std::copy( X, X + dims, ret.X ); return ret; } //umm??? Point< dims, int > operator int() const { Point<dims, int> ret; std::copy( X, X + dims, ret.X ); return ret; } //OK Point<dims, int> toint() { Point<dims, int> ret; std::copy( X, X + dims, ret.X ); return ret; } }; //struct Point template < typename T > Point< 2, T > Create( T X0, T X1 ) { Point< 2, T > ret; ret.X[ 0 ] = X0; ret.X[ 1 ] = X1; return ret; } }; //namespace point int main(void) { using namespace point; Point< 2, double > p2d = point::Create( 12.3, 34.5 ); Point< 2, int > p2i = (int)p2d; //äähhm??? std::cout << p2d.str() << std::endl; char c; std::cin >> c; return 0; } I think the problem is here that C++ cannot distinguish between different return types? many thanks in advance. regards Oops

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  • delegating into private parts

    - by FredOverflow
    Sometimes, C++'s notion of privacy just baffles me :-) class Foo { struct Bar; Bar* p; public: Bar* operator->() const { return p; } }; struct Foo::Bar { void baz() { std::cout << "inside baz\n"; } }; int main() { Foo::Bar b; // error: 'struct Foo::Bar' is private within this context Foo f; f->baz(); // fine } Since Foo::Bar is private, I cannot declare b in main. Yet I can call methods from Foo::Bar just fine. Why the hell is this allowed? Was that an accident or by design?

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  • Detect if class has overloaded function fails on Comeau compiler

    - by Frank
    Hi Everyone, I'm trying to use SFINAE to detect if a class has an overloaded member function that takes a certain type. The code I have seems to work correctly in Visual Studio and GCC, but does not compile using the Comeau online compiler. Here is the code I'm using: #include <stdio.h> //Comeau doesnt' have boost, so define our own enable_if_c template<bool value> struct enable_if_c { typedef void type; }; template<> struct enable_if_c< false > {}; //Class that has the overloaded member function class TestClass { public: void Func(float value) { printf( "%f\n", value ); } void Func(int value) { printf( "%i\n", value ); } }; //Struct to detect if TestClass has an overloaded member function for type T template<typename T> struct HasFunc { template<typename U, void (TestClass::*)( U )> struct SFINAE {}; template<typename U> static char Test(SFINAE<U, &TestClass::Func>*); template<typename U> static int Test(...); static const bool Has = sizeof(Test<T>(0)) == sizeof(char); }; //Use enable_if_c to only allow the function call if TestClass has a valid overload for T template<typename T> typename enable_if_c<HasFunc<T>::Has>::type CallFunc(TestClass &test, T value) { test.Func( value ); } int main() { float value1 = 0.0f; int value2 = 0; TestClass testClass; CallFunc( testClass, value1 ); //Should call TestClass::Func( float ) CallFunc( testClass, value2 ); //Should call TestClass::Func( int ) } The error message is: no instance of function template "CallFunc" matches the argument list. It seems that HasFunc::Has is false for int and float when it should be true. Is this a bug in the Comeau compiler? Am I doing something that's not standard? And if so, what do I need to do to fix it?

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  • does overload operator-> a compile time action?

    - by Brent
    when I tried to compile the code: 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 add "-S" option for compile only.

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  • C++ operator[] syntax.

    - by Lanissum
    Just a quick syntax question. I'm writing a map class (for school). If I define the following operator overload: template<typename Key, typename Val> class Map {... Val* operator[](Key k); What happens when a user writes: Map<int,int> myMap; map[10] = 3; Doing something like that will only overwrite a temporary copy of the [null] pointer at Key k. Is it even possible to do: map[10] = 3; printf("%i\n", map[10]); with the same operator overload?

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