operator-overloading

问题 I am trying to overload the subscript operator -i know it as the element access operator- to take a char * or a std::string. I have a struct struct foo { int Age; const char *Name; }; and a std::vector that will be holding multiple instances of this struct. std::vector<foo>bar; my goal is to be able to access each foo in bar by calling them by their name. std::cout<<"Simon's age is: "<<bar["simon"]; I've just been searching google for a long while trying to find an example or something to go

问题 A while ago, I found in a website some code examples of utility functions that are used when creating , destructing objects, or even when overloading some of their operators . More precisely, the following member functions are mainly used: init, copy, set, and destroy. The init member function is used to initialize all the private members. It's mostly called inside the constructors , e.g. the default or parameter constructor . The copy member function is used to do a deep copy of an object

问题 Foreword In order to store banded matrices whose full counterparts can have both rows and columns indexed from indices other than 1 , I defined a derived data type as TYPE CDS REAL, DIMENSION(:,:), ALLOCATABLE :: matrix INTEGER, DIMENSION(2) :: lb, ub INTEGER :: ld, ud END TYPE CDS where CDS stands for compressed diagonal storage. Given the declaration TYPE(CDS) :: A , The rank-2 component matrix is supposed to contain, as columns, the diagonals of the actual full matrix (like here, except

问题 I have the following code, which is a wrapper for POD into a template class Foo<T> , where T is the wrapped type (can be int , double etc). I define a templated conversion operator, and also the friend addition operator+ , see code below. In the last line of main() , I define Foo<int> a = 10 then compute cout << 2.1 + a << endl; // outputs 12 What is the rule here? It looks like the expression is being translated as operator+(2.1, a) which then becomes operator+(Foo<int>(2.1), a) . Why doesn

问题 I know it is possible to overload operators that already exist in c++ to define desired behavior, but is it possible to create your own operator? Just for example, making an operator # that returns the size of containers: template<typename T> size_t operator#(const T& obj) { return obj.size(); } vector<int> v(1024); cout << #v; // prints 1024 回答1: No. You need to stick with the operators the parser already knows how to handle. Overloading can extend the meaning of expressions, but the syntax

问题 I know it is possible to overload operators that already exist in c++ to define desired behavior, but is it possible to create your own operator? Just for example, making an operator # that returns the size of containers: template<typename T> size_t operator#(const T& obj) { return obj.size(); } vector<int> v(1024); cout << #v; // prints 1024 回答1: No. You need to stick with the operators the parser already knows how to handle. Overloading can extend the meaning of expressions, but the syntax

问题 I am working my way through Barnes' book 'Programming in Ada 2012'. This is a code sample implementing a stack from section 12.5. src/stacks.adb: (the main relevant file) package body Stacks is procedure Push(S: in out Stack; X: in Integer) is begin S := new Cell'(S,X); end Push; procedure Pop(S: in out Stack; X: in out Integer) is begin X := S.Value; S := Stack(S.Next); end Pop; function "="(S, T: Stack) return Boolean is SS: access Cell := S; TT: access Cell := T; begin while SS /= null and

问题 According to 13.3.1.2/8, or better footnote-129 (emphasis mine): [...] The process repeats until an operator-> function returns a value of non-class type . I thought I knew how operator-> works (let me say, its recursive way based on the return type), but I see that I'm completely unaware about how it actually works (I mean, its return type ). When I found it, I wondered if one can really define and use something like a double operator->() for a generic struct S , for I've never used such an

问题 Working on a project I did not initiate, I want to add an << operator to a class. Problem: the class is a private inner class of an other class, the latter being in a namespace . And I cannot make it. The problem can be simplified this way: #include <iostream> #include <map> namespace A { class B { private: typedef std::map<int, int> C; C a; friend std::ostream& operator<<(std::ostream& os, const C &c) { for (C::const_iterator p = c.begin(); p != c.end(); ++p) os << (p->first) << "->" << (p-

问题 For the following generic c# class, I'd like to convert T to K: public abstract class ValueType<T,K> : IValueType<T> where K : ValueType<T,K>,new() { public abstract T Value { get; set; } public static implicit operator ValueType<T,K>(T val) { K k = new K(); k.Value = val; return k; } } If I were to implement a direct operator implicit operator K(T val) it would result in a compile-time error (CS0556). I thought I could try chaining implicit operators: public static implicit operator K