Search Results

Search found 3956 results on 159 pages for 'constructor 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 29/159 | < Previous Page | 25 26 27 28 29 30 31 32 33 34 35 36  | Next Page >

  • How to reduce redundant code when adding new c++0x rvalue reference operator overloads

    - by Inverse
    I am adding new operator overloads to take advantage of c++0x rvalue references, and I feel like I'm producing a lot of redundant code. I have a class, tree, that holds a tree of algebraic operations on double values. Here is an example use case: tree x = 1.23; tree y = 8.19; tree z = (x + y)/67.31 - 3.15*y; ... std::cout << z; // prints "(1.23 + 8.19)/67.31 - 3.15*8.19" For each binary operation (like plus), each side can be either an lvalue tree, rvalue tree, or double. This results in 8 overloads for each binary operation: // core rvalue overloads for plus: tree operator +(const tree& a, const tree& b); tree operator +(const tree& a, tree&& b); tree operator +(tree&& a, const tree& b); tree operator +(tree&& a, tree&& b); // cast and forward cases: tree operator +(const tree& a, double b) { return a + tree(b); } tree operator +(double a, const tree& b) { return tree(a) + b; } tree operator +(tree&& a, double b) { return std::move(a) + tree(b); } tree operator +(double a, tree&& b) { return tree(a) + std::move(b); } // 8 more overloads for minus // 8 more overloads for multiply // 8 more overloads for divide // etc which also has to be repeated in a way for each binary operation (minus, multiply, divide, etc). As you can see, there are really only 4 functions I actually need to write; the other 4 can cast and forward to the core cases. Do you have any suggestions for reducing the size of this code? PS: The class is actually more complex than just a tree of doubles. Reducing copies does dramatically improve performance of my project. So, the rvalue overloads are worthwhile for me, even with the extra code. I have a suspicion that there might be a way to template away the "cast and forward" cases above, but I can't seem to think of anything.

    Read the article

  • How to change the meaning of pointer access operator

    - by kumar_m_kiran
    Hi All, This may be very obvious question, pardon me if so. I have below code snippet out of my project, #include <stdio.h> class X { public: int i; X() : i(0) {}; }; int main(int argc,char *arv[]) { X *ptr = new X[10]; unsigned index = 5; cout<<ptr[index].i<<endl; return 0; } Question Can I change the meaning of the ptr[index] ? Because I need to return the value of ptr[a[index]] where a is an array for subindexing. I do not want to modify existing source code. Any new function added which can change the behavior is needed. Since the access to index operator is in too many places (536 to be precise) in my code, and has complex formulas inside the index subscript operator, I am not inclined to change the code in many locations. PS : 1. I tried operator overload and came to conclusion that it is not possible. 2. Also p[i] will be transformed into *(p+i). I cannot redefine the basic operator '+'. So just want to reconfirm my understanding and if there are any possible short-cuts to achieve. Else I need fix it by royal method of changing every line of code :) .

    Read the article

  • Implicit array casting in C#

    - by Malki
    Hi, I have the following classes with an implicit cast operator defined: class A { ... } class B { private A m_a; public B(A a) { this.m_a = a; } public static implicit operator B(A a) { return new B(a); } } Now, I can implicitly cast A to B. But why can't I implicitly cast A[] to B[] ? static void Main(string[] args) { // compiles A a = new A(); B b = a; // doesn't compile A[] arrA = new A[] {new A(), new A()}; B[] arrB = arrA; } Thanks, Malki.

    Read the article

  • 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.

    Read the article

  • How do I write an overload operator where both arguments are interface

    - by Eric Girard
    I'm using interface for most of my stuff. I can't find a way to create an overload operator + that would allow me to perform an addition on any objects implementing the IPoint interface Code interface IPoint { double X { get; set; } double Y { get; set; } } class Point : IPoint { double X { get; set; } double Y { get; set; } //How and where do I create this operator/extension ??? public static IPoint operator + (IPoint a,IPoint b) { return Add(a,b); } public static IPoint Add(IPoint a,IPoint b) { return new Point { X = a.X + b.X, Y = a.Y + b.Y }; } } //Dumb use case : public class Test { IPoint _currentLocation; public Test(IPoint initialLocation) { _currentLocation = intialLocation } public MoveOf(IPoint movement) { _currentLocation = _currentLocation + intialLocation; //Much cleaner/user-friendly than _currentLocation = Point.Add(_currentLocation,intialLocation); } }

    Read the article

  • 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).

    Read the article

  • 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.

    Read the article

  • Invoking a method overloaded where all arguments implement the same interface

    - by double07
    Hello, My starting point is the following: - I have a method, transform, which I overloaded to behave differently depending on the type of arguments that are passed in (see transform(A a1, A a2) and transform(A a1, B b) in my example below) - All these arguments implement the same interface, X I would like to apply that transform method on various objects all implementing the X interface. What I came up with was to implement transform(X x1, X x2), which checks for the instance of each object before applying the relevant variant of my transform. Though it works, the code seems ugly and I am also concerned of the performance overhead for evaluating these various instanceof and casting. Is that transform the best I can do in Java or is there a more elegant and/or efficient way of achieving the same behavior? Below is a trivial, working example printing out BA. I am looking for examples on how to improve that code. In my real code, I have naturally more implementations of 'transform' and none are trivial like below. public class A implements X { } public class B implements X { } interface X { } public A transform(A a1, A a2) { System.out.print("A"); return a2; } public A transform(A a1, B b) { System.out.print("B"); return a1; } // Isn't there something better than the code below??? public X transform(X x1, X x2) { if ((x1 instanceof A) && (x2 instanceof A)) { return transform((A) x1, (A) x2); } else if ((x1 instanceof A) && (x2 instanceof B)) { return transform((A) x1, (B) x2); } else { throw new RuntimeException("Transform not implemented for " + x1.getClass() + "," + x2.getClass()); } } @Test public void trivial() { X x1 = new A(); X x2 = new B(); X result = transform(x1, x2); transform(x1, result); }

    Read the article

  • Cant overload python socket.send

    - by ralu
    Code from socket import socket class PolySocket(socket): def __init__(self,*p): print "PolySocket init" socket.__init__(self,*p) def sendall(self,*p): print "PolySocket sendall" return socket.sendall(self,*p) def send(self,*p): print "PolySocket send" return socket.send(self,*p) def connect(self,*p): print "connecting..." socket.connect(self,*p) print "connected" HOST="stackoverflow.com" PORT=80 readbuffer="" s=PolySocket() s.connect((HOST, PORT)) s.send("a") s.sendall("a") Output: PolySocket init connecting... connected PolySocket sendall As we can see, send method is not overloaded.

    Read the article

  • 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.

    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

  • 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

  • [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

  • 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

  • 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

  • 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

  • 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

  • Difficulty creating classes and arrays of those classes C#

    - by Lucifer Fayte
    I'm trying to implement a Discrete Fourier Transformation algorithm for a project I'm doing in school. But creating a class is seeming to be difficult(which it shouldn't be). I'm using Visual Studio 2012. Basically I need a class called Complex to store the two values I get from a DFT; The real portion and the imaginary portion. This is what I have so far for that: using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace SoundEditor_V3 { public class Complex { public double real; public double im; public Complex() { real = 0; im = 0; } } } The problem is that it doesn't recognize the constructor as a constructor, now I'm just learning C#, but I looked it up online and this is how it's supposed to look apparently. It recognizes my constructor as a method. Why is that? Am I creating the class wrong? It's doing the same thing for my Fourier class as well. So each time I try to create a Fourier object and then use it's method...there is no such thing. example, I do this: Fourier fou = new Fourier(); fou.DFT(s, N, amp, 0); and it tells me fou is a 'field' but is used like a 'type' why is it saying that? Here is the code for my Fourier class as well: using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace SoundEditor_V3 { public class Fourier { //FOURIER //N = number of samples //s is the array of samples(data) //amp is the array where the complex result will be written to //start is the where in the array to start public void DFT(byte[] s, int N, ref Complex[] amp, int start) { Complex tem = new Complex(); int f; int t; for (f = 0; f < N; f++) { tem.real = 0; tem.im = 0; for (t = 0; t < N; t++) { tem.real += s[t + start] * Math.Cos(2 * Math.PI * t * f / N); tem.im -= s[t + start] * Math.Sin(2 * Math.PI * t * f / N); } amp[f].real = tem.real; amp[f].im = tem.im; } } //INVERSE FOURIER public void IDFT(Complex[] A, ref int[] s) { int N = A.Length; int t, f; double result; for (t = 0; t < N; t++) { result = 0; for (f = 0; f < N; f++) { result += A[f].real * Math.Cos(2 * Math.PI * t * f / N) - A[f].im * Math.Sin(2 * Math.PI * t * f / N); } s[t] = (int)Math.Round(result); } } } } I'm very much stuck at the moment, any and all help would be appreciated. Thank you.

    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

  • 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

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 | 25 26 27 28 29 30 31 32 33 34 35 36  | Next Page >