rvalue

问题 The C++ standard defines the following functions deleted; template <class T> void ref(const T&&) = delete; template <class T> void cref(const T&&) = delete; This is to aid in ensuring that the functions are not misused by disallowing them from binding to temporary values (rvalues). Does const && bind to all rvalues, specifically prvalues? Would const && bind to all "moved objects" (xvalues; basically something returned from std::move or similar)? I can reason that it should, but I don't have

问题 The C++ standard defines the following functions deleted; template <class T> void ref(const T&&) = delete; template <class T> void cref(const T&&) = delete; This is to aid in ensuring that the functions are not misused by disallowing them from binding to temporary values (rvalues). Does const && bind to all rvalues, specifically prvalues? Would const && bind to all "moved objects" (xvalues; basically something returned from std::move or similar)? I can reason that it should, but I don't have

问题 The Rust Reference says: The left operand of an assignment or compound-assignment expression is an lvalue context, as is the single operand of a unary borrow. [...] When an rvalue is used in an lvalue context, a temporary un-named lvalue is created and used instead. This rvalue promotion obviously works with borrowing: let ref_to_i32 = &27; // a temporary i32 variable with value 27 is created But it doesn't seem to work in an assignment (although the reference speaks about all lvalue contexts

问题 The Rust Reference says: The left operand of an assignment or compound-assignment expression is an lvalue context, as is the single operand of a unary borrow. [...] When an rvalue is used in an lvalue context, a temporary un-named lvalue is created and used instead. This rvalue promotion obviously works with borrowing: let ref_to_i32 = &27; // a temporary i32 variable with value 27 is created But it doesn't seem to work in an assignment (although the reference speaks about all lvalue contexts

问题 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

问题 C++03 §4.2 N°1: An lvalue or rvalue of type “array of N T” or “array of unknown bound of T” can be converted to an rvalue of type “pointer to T.” The result is a pointer to the first element of the array. What has been confusing in this statement for a long time for me was that I didn't quite understand what an rvalue of array type would mean. That is, I couldn't come up with an expression whose type were an array and the result were an rvalue. I read this thread, which basically asks the

问题 In C++, pre-increment operator gives lvalue because incremented object itself is returned, not a copy. But in C, it gives rvalue. Why? 回答1: 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’’ <=> rvalue However for

问题 This question already has answers here : Literal initialization for const references (3 answers) Closed 5 years ago . We cannot write int& ref = 40 because we need lvalue on right side. But we can write const int& ref = 40 . Why is this possible? 40 is rvalue instead lvalue I know that this is an exception but why? 回答1: As Stroustrup says: The initializer for a const T& need not be an lvalue or even of type T. In such cases: [1] First, implicit type conversion to T is applied if necessary. [2

问题 I've been reading quite many on the Internet and it seems that many people mentioned the following rules (but i couldn't find it in the standard), The addition operator + (and all other binary operators) requires both operands to be rvalue, and the result is rvalue. And so on.. I checked the C++ standard, and it clearly states that (clause 3.10/2), Whenever a glvalue appears in a context where a prvalue is expected, the glvalue is converted to a prvalue (clause 5/9), Whenever a glvalue

问题 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