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  • How operator oveloading works

    - by Rasmi Ranjan Nayak
    I have below code class rectangle { ..... .....//Some code int operator+(rectangle r1) { return(r1.length+length); } }; In main fun. int main() { rectangle r1(10,20); rectangle r2(40,60); rectangle r3(30,60); int len = r1+r3; } Here if we will see in operator+(), we are doing r1.length + length. How the compiler comes to know that the 2nd length in return statement belong to object r3 not to r1 or r2? I think answer may be in main() we have writeen int len = r1+r3; If that is the case then why do we need to write in operator+(....) { r1.lenth + lenth; //Why not length + length? } Why not length + length? Bcause compiler already knows from main() that the first length belong to object r1 and 2nd to object r3.

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  • What are the default return values for operator< and operator[] in C++ (Visual Studio 6)?

    - by DustOff
    I've inherited a large Visual Studio 6 C++ project that needs to be translated for VS2005. Some of the classes defined operator< and operator[], but don't specify return types in the declarations. VS6 allows this, but not VS2005. I am aware that the C standard specifies that the default return type for normal functions is int, and I assumed VS6 might have been following that, but would this apply to C++ operators as well? Or could VS6 figure out the return type on its own? For example, the code defines a custom string class like this: class String { char arr[16]; public: operator<(const String& other) { return something1 < something2; } operator[](int index) { return arr[index]; } }; Would VS6 have simply put the return types for both as int, or would it have been smart enough to figure out that operator[] should return a char and operator< should return a bool (and not convert both results to int all the time)? Of course I have to add return types to make this code VS2005 C++ compliant, but I want to make sure to specify the same type as before, as to not immediately change program behavior (we're going for compatibility at the moment; we'll standardize things later).

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  • 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.

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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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  • [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.

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  • No-argument method on window.external is invoked when checking with typeof

    - by janko
    Hi, I am trying to display an HTML page with embedded JavaScript code inside a System.Windows.Forms.WebBrowser control. The JavaScript code is expected to interact with the embedding environment through the window.external object. Before invoking a method on window.external, JavaScript is supposed to check for the existance of the method. If it is not there, the code should invoke a generic fallback method. // basic idea if (typeof(window.external.MyMethod) != 'undefined') { window.external.MyMethod(args); } else { window.external.Generic("MyMethod", args); } However, checking for a no-argument method with typeof seems to invoke the method already. That is, if MyMethod accepts any positive number of arguments, the code above will work perfectly; but, if MyMethod is a no-argument method, then the expression typeof(window.external.MyMethod) will not check for its type but invoke it, too. Is there any work-around to this behavior? Can I somehow escape the expression window.external.MyMethod to prevent the method call from occurring?

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  • 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); ?

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  • 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?

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  • 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

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  • 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 } }

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  • 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];

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  • 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; }

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  • 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?

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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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  • 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 !

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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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  • 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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  • 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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