I am deriving stl container class vector and trying to overload the [] operators to provide range checking on accessing element with subscript. But I get compile errors when try to create const object of my derived class Vec . For example below is the code: template<class T> class Vec : public vector<T> { public: Vec(): vector<T>() { } Vec(int i):vector&l
我正在派生stl容器类矢量并试图重载[]运算符来提供使用下标访问元素的范围检查。 但是当我尝试创建派生类Vec const对象时，我得到编译错误。 例如下面是代码： template<class T> class Vec : public vector<T> { public: Vec(): vector<T>() { } Vec(int i):vector<T>(i) { } T& operator[](int i) { return at(i); } const T& operat
I am learning c++ on my own. I was studying operator overloading, i was able to understand addition and subtraction operator overloading. But overloading of I/O operators is a bit confusing. I have created a class for Complex numbers, now i am overloading operators. Function prototype from Complex.h friend ostream& operator<<(ostream&, const Complex&); Function from Compl
我自己学习c ++。 我正在研究运算符重载，我能够理解加减运算符的重载。 但I / O操作员的过载有点令人困惑。 我创建了复杂数字的类，现在我正在重载运算符。 Complex.h中的函数原型 friend ostream& operator<<(ostream&, const Complex&); 来自Complex.cpp的函数 ostream& operator<<(ostream& os, const Complex& value){ os << "(" << value.r <<", " <<
I am working to recreate the various c++ types in order to better understand how they work. I am currently stuck on the += operator and can not find what the issue is with my declaration. Here is the code for my Class: class String { int size; char * buffer; public: String(); String(const String &); String(const char *); int length(){return size;}; friend bool
我正在努力重新创建各种c ++类型，以便更好地理解它们的工作方式。 我目前停留在+ =运算符上，无法找到我的声明出现的问题。 以下是我的课程代码： class String { int size; char * buffer; public: String(); String(const String &); String(const char *); int length(){return size;}; friend bool operator==(const String &, const String &); friend bool operator<=
I'm trying to overload the << operator but I'm getting some errors of this kind: passing const std::ostream' as this' argument of `std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(const void*) [with _CharT = char, _Traits = std::char_traits]' discards qualifiers Here is my code: #include<iostream> using
我试图重载<<运算符，但我收到了这样的错误： 传递const std::ostream' as此参数的std :: basic_ostream <_CharT，_Traits>＆std :: basic_ostream <_CharT，_Traits> ::运算符<<（const void *）[with _CharT = char，_Traits = std :: char_traits]'丢弃限定符 这是我的代码： #include<iostream> using namespace std; class nod{ protected: int info; nod *next;
This is my standalone operator overloading function. #include <fstream> #include <iostream> #include <sstream> #include <string> #include <vector> using namespace std; typedef vector <string> record_t; typedef vector <record_t> data_t; istream& operator >> ( istream& ins, record_t& record ) { record.clear(); string line; getl
这是我独立的运算符重载功能。 #include <fstream> #include <iostream> #include <sstream> #include <string> #include <vector> using namespace std; typedef vector <string> record_t; typedef vector <record_t> data_t; istream& operator >> ( istream& ins, record_t& record ) { record.clear(); string line; getline( ins, line ); stri
Code: std::ostream& operator<<(std::ostream& os, const BmvMessage& bm); I don't see anything incorrect, but it gives the following error: error: `std::ostream& BMV::BmvMessage::operator<<(std::ostream&, const BMV::BmvMessage&)' must take exactly one argument. I do not know why this happens. Any suggestions are welcome. I have done this before an
码： std::ostream& operator<<(std::ostream& os, const BmvMessage& bm); 我没有看到任何不正确的东西，但它给出了以下错误： 错误：`std :: ostream＆BMV :: BmvMessage :: operator <<（std :: ostream＆，const BMV :: BmvMessage＆）'必须采用恰好一个参数。 我不知道为什么会发生这种情况。 欢迎任何建议。 我以前做过这个，从来没有遇到过这个错误。 我也在网上查过，它看起来像：
I am writing a small matrix library in C++ for matrix operations. However my compiler complains, where before it did not. This code was left on a shelf for 6 months and in between I upgraded my computer from debian etch to lenny (g++ (Debian 4.3.2-1.1) 4.3.2 ) however I have the same problem on a Ubuntu system with the same g++. Here is the relevant part of my matrix class: namespace Math {
我正在用C ++编写一个矩阵运算的小型矩阵库。 然而，我的编译器抱怨，之前没有。 这段代码放置在书架上6个月，之间我将计算机从debian etch升级到lenny（g ++（Debian 4.3.2-1.1）4.3.2），但是我在Ubuntu系统上使用相同的g ++ 。 这是我矩阵类的相关部分： namespace Math { class Matrix { public: [...] friend std::ostream& operator<< (std::ostream& stream, const Matr
Is it possible to overload [] operator twice? To allow, something like this: function[3][3] (like in a two dimensional array). If it is possible, I would like to see some example code. You can overload operator[] to return an object on which you can use operator[] again to get a result. class ArrayOfArrays { public: ArrayOfArrays() { _arrayofarrays = new int*[10]; for(in
是否可以重载[]运算符两次？ 要允许，像这样： function[3][3] （就像在一个二维数组中）。 如果可能的话，我希望看到一些示例代码。 你可以重载operator[]来返回一个可以再次使用operator[]的对象来获得结果。 class ArrayOfArrays { public: ArrayOfArrays() { _arrayofarrays = new int*[10]; for(int i = 0; i < 10; ++i) _arrayofarrays[i] = new int[10]; } class Proxy
I am trying to overload the friend >> operator with templates. I don't want to define it inline. I had tried to do the same with the help of the method add() defined in the code below. It works fine. I would like my >> operator to do the same. The following is my code: #include<iostream> template<class T>class Demo; template<class T> std::ostream& op
我想用模板超载朋友>>操作符。 我不想在内联中定义它。 我试图在下面的代码中定义的方法add()的帮助下做同样的事情。 它工作正常。 我希望我的>>操作员也这样做。 以下是我的代码： #include<iostream> template<class T>class Demo; template<class T> std::ostream& operator<<(std::ostream&, const Demo<T> &); template<class T> std::istream& o
Learning c++ bit by bit in codeblocks, I encountered this error. I have tried it for hours and saw various links but its still not working, main.cpp #include <iostream> #include "Person.h" using namespace std; int main() { foo:: Person p; //std:: cin >> p; std:: cout << p; return 0; } Person.h #ifndef PERSON_H #define PERSON_H namespace foo{ class Person {
在代码块中逐位学习c ++，我遇到了这个错误。 我已经尝试了几个小时，看到各种链接，但它仍然无法正常工作， main.cpp中 #include <iostream> #include "Person.h" using namespace std; int main() { foo:: Person p; //std:: cin >> p; std:: cout << p; return 0; } Person.h #ifndef PERSON_H #define PERSON_H namespace foo{ class Person { public: int age; Person();