Are objects inside rvalue referenced object, also rvalue referenced?
struct A{
};
struct B{
A a2;
};
//template<class B>
void test(B &&b){
// 1. Is this the correct way?
auto &&in3 = std::forward<B>(b).a2;
std::cout << std::is_rvalue_reference<decltype(in3)>::value;
// return true
// 2. or this?
auto &&in4 = b.a2;
std::cout << std::is_rvalue_reference<decltype(in4)>::value;
// return false
}
test(B());
http://coliru.stacked-crooked.com/a/bcf0f7dc4cc0440e
An rvalue reference is formed by placing an && after some type. An rvalue reference behaves just like an lvalue reference except that it can bind to a temporary (an rvalue), whereas you can not bind a (non const) lvalue reference to an rvalue.
Explanation: If you pass an lvalue T to enqueue , U will deduce to T& , and the forward will pass it along as an lvalue, and you'll get the copy behavior you want. If you pass an rvalue T to enqueue , U will deduce to T , and the forward will pass it along as an rvalue, and you'll get the move behavior you want.
“l-value” refers to a memory location that identifies an object. “r-value” refers to the data value that is stored at some address in memory. References in C++ are nothing but the alternative to the already existing variable. They are declared using the '&' before the name of the variable.
2) non-modifiable lvalues, which are const. rvalue — The expression that refers to a disposable temporary object so they can't be manipulated at the place they are created and are soon to be destroyed. An address can not be taken of rvalues. An rvalue has no name as its a temporary value.
Yes, members of rvalues are themselves rvalues. This was clarified by DR 421
But that is irrelevant here:
auto &&in4 = b.a2;
b
is not an rvalue, it's an lvalue (simple rule of thumb: it has a name).
To restore the value category it had when passed to the function you need to forward
it
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