rvalue

void test(int && val) { val=4; } void main() { test(1); std::cin.ignore(); } Is a int is created when test is called or by default in c++ literals are int type? Nawaz Note that your code would compile only with C++11 compiler. When you pass an integral literal, which is by default of int type, unless you write 1L , a temporary object of type int is created which is bound to the parameter of the function. It's like the first from the following initializations: int && x = 1; //ok. valid in C++11 only. int & y = 1; //error, both in C++03, and C++11 const int & z = 1; //ok, both in C++03, and C+

It took me quite some time to understand the difference between an rvalue and a temporary object. But now the final committee draft states on page 75: An rvalue [...] is an xvalue, a temporary object or subobject thereof, or a value that is not associated with an object. I can't believe my eyes. This must be an error, right? To clarify, here is how I understand the terms: #include <string> void foo(std::string&& str) { std::cout << str << std::endl; } int main() { foo(std::string("hello")); } In this program, there are two expressions that denote the same temporary object : the prvalue std:

Why don't rvalues have a memory address? Are they not loaded into the RAM when the program executes or does it refer to the values stored in processor registers? Your question ("Why don't rvalues have a memory address?") is a bit confused. An rvalue is a kind of expression. Expressions don't have addresses: objects have addresses. It would be more correct to ask "why can one not apply the address-of operator to an rvalue expression?" The answer to that is rather simple: you can only take the address of an object and not all rvalue expressions refer to objects (for example, the expression 42

I have a View and a Shape class where the View "owns" its Shape objects. I am implementing this as a vector of unique_ptr. In the function View::add_shape(std::unique_ptr&& shape), I still need to use std::move on the rvalue parameter to make it compile. Why? ( using GCC 4.8 ) #include <memory> #include <vector> using namespace std; class Shape { }; class View { vector<unique_ptr<Shape>> m_shapes; public: void add_shape(unique_ptr<Shape>&& shape) { m_shapes.push_back(std::move(shape));// won't compile without the std::move } }; int main() { unique_ptr<Shape> ups(new Shape); View v; v.add_shape

Considering the following code: #include <iostream> using namespace std; struct I { I(I&& rv) { cout << "I::mvcotr" << endl; } }; struct C { I i; I&& foo() { return move(i) }; } }; int main() { C c; I i = c.foo(); } C contains I. And C::foo() allows you to move I out of C. What is the difference between the member function used above: I&& foo() { return move(i) }; // return rvalue ref and the following replacement member function: I foo() { return move(i) }; // return by value To me, they seem to do the same thing: I i = c.foo(); leads to a call to I::I(I&&); . What consequences will there be

The implementation of std::move basically looks like this: template<typename T> typename std::remove_reference<T>::type&& move(T&& t) { return static_cast<typename std::remove_reference<T>::type&&>(t); } Note that the parameter of std::move is a universal reference (also known as a forwarding reference, but we're not forwarding here). That is, you can std::move both lvalues and rvalues: std::string a, b, c; // ... foo(std::move(a)); // fine, a is an lvalue foo(std::move(b + c)); // nonsense, b + c is already an rvalue But since the whole point of std::move is to cast to an rvalue, why are we

The code is as follows: #include <iostream> using namespace std; class A { }; A rtByValue() { return A(); } void passByRef(A &aRef) { // do nothing } int main() { A aa; rtByValue() = aa; // compile without errors passByRef(rtByValue()); // compile with error return 0; } The g++ compiler gives the following error: d.cpp: In function ‘int main()’: d.cpp:19:23: error: invalid initialization of non-const reference of type ‘A&’ from an rvalue of type ‘A’ d.cpp:12:6: error: in passing argument 1 of ‘void passByRef(A&)’ It says that I can't pass an rvalue as an argument of a non-const reference, but