I\'ve been studying rvalue references lately and came to a conclusion that it\'s quite advantageous t开发者_如何学运维o use pass-by-value everywhere where complete copy of an object will be made (for
Consider the following function: Foo foo(Foo x) { return x; } Will return x invoke the copy constructor or the move constructor? (Let\'s leave NRVO aside here.)
Is it possible to design and how should I make overloaded operator+ for my class C to have this possible:
I am wondering if in C++0x \"12.8 C开发者_如何学Copying and Moving class objects [class.copy] paragraph 31\" when copy elision happens, exactly:
Is there a reason when a function should return a RValue Reference? A t开发者_开发技巧echnique, or trick, or an idiom or pattern?
Is the following snipplet correct for un-defining all otherwise generated methods and constructors for a class?
#include <type_traits> template<class开发者_StackOverflow中文版 T> typename std::remove_reference<T>::type&& move(T&& v)
If I define a function which accepts an rvalue reference parameter: template <typename T> void fooT(T &&x) {}
I was viewing the MSDN doc about mu开发者_运维问答ltimap and find that it has a member function multimap::emplace(). Below is the example of that member function.
I\'ve been looking into some of the new features of C++11 and one I\'ve noticed is the double ampersand in declaring variables, like T&& var.
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