operator-keyword

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? aschepler This is C++11 code. In C++11, the && token can be used to mean an "rvalue reference". Lexseal Lin && is new in C++11. int&& a means "a" is an r-value reference. && is normally only used to declare a parameter of a function

I'm having trouble with the inheritance of operator=. Why doesn't this code work, and what is the best way to fix it? #include <iostream> class A { public: A & operator=(const A & a) { x = a.x; return *this; } bool operator==(const A & a) { return x == a.x; } virtual int get() = 0; // Abstract protected: int x; }; class B : public A { public: B(int x) { this->x = x; } int get() { return x; } }; class C : public A { public: C(int x) { this->x = x; } int get() { return x; } }; int main() { B b(3); C c(7); printf("B: %d C: %d B==C: %d\n", b.get(), c.get(), b==c); b = c; // compile error // error:

Why isn't operator.iadd(x, y) equivalent to z = x; z += y ? And how does operator.iadd(x, y) differ from operator.add(x, y) ? From the docs : Many operations have an “in-place” version. The following functions provide a more primitive access to in-place operators than the usual syntax does; for example, the statement x += y is equivalent to x = operator.iadd(x, y). Another way to put it is to say that z = operator.iadd(x, y) is equivalent to the compound statement z = x; z += y. Related question , but I am not interested in Python class methods; just regular operators on built-in Python types.