operator-keyword

问题 I want to implement "operator * " overloading INSIDE my class, so I would be able to do the following: Rational a(1, 2), b; b = 0.5 * a; // b = 1/4 Notice that b is on the right side, is there a way to do such a thing inside "Rational" class? 回答1: No. You must define operator* as a free function. Of course, you could implement it in terms of a member function on the second argument. 回答2: Yes: class Rational { // ... friend Rational operator*(float lhs, Rational rhs) { rhs *= lhs; return rhs;

问题 XML document: <doc> <A> <Node>Hello!</Node> </A> <B> <Node/> </B> <C> </C> <D/> </doc> How would you evaluate the following XPath queries? /doc/A/Node != 'abcd' /doc/B/Node != 'abcd' /doc/C/Node != 'abcd' /doc/D/Node != 'abcd' I would expect ALL of these to evaluate to true . However, here are the results: /doc/A/Node != 'abcd' true /doc/B/Node != 'abcd' true /doc/C/Node != 'abcd' false /doc/D/Node != 'abcd' false Is this expected behavior? Or is it a bug with my XPath provider (jaxen)? 回答1:

问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: How do I check for nulls in an ‘==’ operator overload without infinite recursion? There is probably an easy answer to this...but it seems to be eluding me. Here is a simplified example: public class Person { public string SocialSecurityNumber; public string FirstName; public string LastName; } Let's say that for this particular application, it is valid to say that if the social security numbers match, and both

问题 This simple program (when compiled on Linux) will CORRECTLY give two different answers based on whether it's compiled with -std=c++0x or not. Problem: I cannot reproduce the same thing on OS X (Mountain Lion, 10.8 SDK). What am I missing? #include <iostream> #include <sstream> class Thing : public std::ostringstream { public: Thing() : std::ostringstream() {} virtual ~Thing() { std::cerr << str(); } }; int main(int argc, const char * argv[]) { Thing() << "Hello" << std::endl; return 0; } To

问题 the following script with debug option 'set -e -v' fails at the increment operator only when the variable has a prior value of zero. #!/bin/bash set -e -v i=1; let i++; echo "I am still here" i=0; let i++; echo "I am still here" i=0; ((i++)); echo "I am still here" bash (GNU bash, version 4.0.33(1)-release (x86_64-apple-darwin10) but also GNU bash, version 4.2.4(1)-release (x86_64-unknown-linux-gnu)) any ideas? 回答1: the answer to my question is not to use let (or shift , or...) but to use i=$

问题 I have difficulties in understanding the sequence of calls in the code below. I was expecting to see the output below A1B2 While I can see that the output I get is BA12 I thought that the call std::cout<< b->fooA() << b->fooB() << std::endl was equivalent to call std::cout.operator<<( b->fooA() ).operator<< ( b->fooB() ) but I can see that this is not the case. Can you help me understanding better how this does it work and the relationship with the global operator<< ? Is this last ever called

问题 I have class CMatrix, where is "double pointer" to array of values. class CMatrix { public: int rows, cols; int **arr; }; I simply need to access the values of matrix by typing: CMatrix x; x[0][0] = 23; I know how to do that using: x(0,0) = 23; But I really need to do that the other way. Can anyone help me with that? Please? Thank you guys for help at the end I did it this way... class CMatrix { public: int rows, cols; int **arr; public: int const* operator[]( int const y ) const { return

问题 Can I define in C++ an array operator that takes multiple arguments? I tried it like this: const T& operator[](const int i, const int j, const int k) const{ return m_cells[k*m_resSqr+j*m_res+i]; } T& operator[](const int i, const int j, const int k){ return m_cells[k*m_resSqr+j*m_res+i]; } But I'm getting this error: error C2804 binary operator '[' has too many parameters 回答1: Nope, you can't overload operator[] to accept multiple arguments. You instead can overload operator() . See How do I

问题 I'm reading STL source code and I have no idea what && address operator is supposed to do. Here is a code example from stl_vector.h : vector& operator=(vector&& __x) // <-- Note double ampersands here { // NB: DR 675. this->clear(); this->swap(__x); return *this; } Does "Address of Address" make any sense? Why does it have two address operators instead of just one? 回答1: This is C++11 code. In C++11, the && token can be used to mean an "rvalue reference". 回答2: && is new in C++11. int&& a means

问题 I'm reading STL source code and I have no idea what && address operator is supposed to do. Here is a code example from stl_vector.h : vector& operator=(vector&& __x) // <-- Note double ampersands here { // NB: DR 675. this->clear(); this->swap(__x); return *this; } Does "Address of Address" make any sense? Why does it have two address operators instead of just one? 回答1: This is C++11 code. In C++11, the && token can be used to mean an "rvalue reference". 回答2: && is new in C++11. int&& a means