move-constructor

I want to inherit copy constructor of the base class using using keyword: #include <iostream> struct A { A() = default; A(const A &) { std::cerr << __PRETTY_FUNCTION__ << std::endl; } A( A &&) { std::cerr << __PRETTY_FUNCTION__ << std::endl; } A& operator=(const A &) { std::cerr << __PRETTY_FUNCTION__ << std::endl; return *this; } A& operator=( A &&) { std::cerr << __PRETTY_FUNCTION__ << std::endl; return *this; } }; struct B : A { using A::A; using A::operator=; B& operator=(const B &) { std::cerr << __PRETTY_FUNCTION__ << std::endl; return *this; } B& operator=( B &&) { std::cerr << __PRETTY

From this reference, it allows a const rvalue as a move constructor Type::Type( const Type&& other ); How can a movable object be const ? Even if this was technically allowed, is there a case where such declaration would be useful? Jonathan Wakely How can a movable object be const ? It can't, but that's not what the language says. The language says that a constructor with that signature is a "move constructor" but that doesn't mean the argument gets moved from, it just means the constructor meets the requirements of a "move constructor". A move constructor is not required to move anything, and

This question already has an answer here: Rvalue Reference is Treated as an Lvalue? 4 answers Lvalue reference constructor is called instead of rvalue reference constructor 1 answer Say I have the following function void doWork(Widget && param) // param is an LVALUE of RRef type { Widget store = std::move(param); } Why do I need to cast param back to an rvalue with std::move() ? Shouldn't it be obvious that the type of param is rvalue since it was declared in the function signature as an rvalue reference? Shouldn't the move constructor be automatically invoked here on this principle alone? Why

The below code can be compiled successfully using Visual Studio 2015, but it failed using Visual Studio 2017. Visual Studio 2017 reports: error C2280: “std::pair::pair(const std::pair &)”: attempting to reference a deleted function Code #include <unordered_map> #include <memory> struct Node { std::unordered_map<int, std::unique_ptr<int>> map_; // Uncommenting the following two lines will pass Visual Studio 2017 compilation //Node(Node&& o) = default; //Node() = default; }; int main() { std::vector<Node> vec; Node node; vec.push_back(std::move(node)); return 0; } It looks like Visual Studio

The first example: #include <iostream> #include <memory> using namespace std; struct A { unique_ptr<int> ref; A(const A&) = delete; A(A&&) = default; A(const int i) : ref(new int(i)) { } ~A() = default; }; int main() { A a[2] = { 0, 1 }; return 0; } It works perfectly. So here the MOVE constructor is used. Let's remove the move constructor and add a copy one: #include <iostream> #include <memory> using namespace std; struct A { unique_ptr<int> ref; A(const A&a) : ref( a.ref.get() ? new int(*a.ref) : nullptr ) { } A(A&&) = delete; A(const int i) : ref(new int(i)) { } ~A() = default; }; int main

I've been testing some C++11 features from some some. I came across r-value references and move constructors. I implemented my first move constructor, here it is: #include <iostream> #include <vector> using namespace std; class TestClass{ public: TestClass(int s): size(s), arr(new int[s]){ } ~TestClass(){ if (arr) delete arr; } // copy constructor TestClass(const TestClass& other): size(other.size), arr(new int[other.size]){ std::copy(other.arr, other.arr + other.size, arr); } // move constructor TestClass(TestClass&& other){ arr=other.arr; size=other.size; other.arr=nullptr; other.size=0; }