A member of an rvalue expression is an rvalue. The standard states in 5.3.5 [expr.ref]: If E2 is declared to have type “reference to T”, then E1.E2 is an lvalue […] – If E2 is a non-static data member, and the type of E1 is “cq1 vq1 X”, and the type of E2 is “cq2 … Read more
C doesn’t have references. In C++ ++i returns a reference to i (lvalue) whereas in C it returns a copy(incremented). C99 6.5.3.1/2 The value of the operand of the prefix ++ operator is incremented. The result is the new value of the operand after incrementation. The expression ++Eis equivalent to (E+=1). ‘‘value of an expression’’ … Read more
string literals are lvalues, but you can’t change them rvalue, but if it’s a pointer and non-NULL, the object it points to is an lvalue The C standard recognizes the original terms stood for left and right as in L = R; however, it says to think of lvalue as locator value, which roughly means … Read more
The committee already seems to be aware that there’s a problem in this part of the standard. CWG issue 690 talks about a somewhat similar problem with exactly the same part of the standard (in the “additional note” from September, 2009). I’d guess new language will be drafted for that part of the standard soon. … Read more
The reason this fails is because when you specify foo<Movable>, the function you’re binding to is: void foo(Movable&&) // *must* be an rvalue { } However, the value passed by std::bind will not be an rvalue, but an lvalue (stored as a member somewhere in the resulting bind functor). That, is the generated functor is … Read more
The word “shall” in the standard language means a strict requirement. So, yes, your code is ill-formed (it is an error) because it attempts to apply address-of operator to a non-lvalue. However, the problem here is not an attempt of taking address of a temporary. The problem is, again, taking address of a non-lvalue. Temporary … Read more
No. The C++ language specification never makes such a straightforward assertion as the one you are asking about. It doesn’t say anywhere in the language standard that “all temporary objects are rvalues”. Moreover, the question itself is a bit of misnomer, since the property of being an rvalue in the C++ language is not a … Read more
Is bar an rvalue or an lvalue? The question answers itself. Whatever has a name is an lvalue(1). So bar is an lvalue. Its type is “rvalue reference to string“, but it’s an lvalue of that type. If you want to treat it as an rvalue, you need to apply std::move() to it. If you … Read more
Do rvalue references allow dangling references? If you meant “Is it possible to create dangling rvalue references” then the answer is yes. Your example, however, string middle_name () { return “Jaan”; } int main() { string&& nodanger = middle_name(); // OK. // The life-time of the temporary is extended // to the life-time of the … Read more
I think it happened the other way around. The initial idea was to introduce rvalue-references into the language, meaning that “the code providing the double-ampersand reference does not care about what will happen to the referred-to object”. This permits move semantics. This is nice. Now. The standard forbids constructing a reference to a reference, but … Read more