move-semantics

问题 I am using Visual Studio 2012 Update 2 and am having trouble trying to understand why std::vector is trying to use the copy constructor of unique_ptr. I have looked at similar issues and most are related to not having an explicit move constructor and/or operator. If I change the member variable to a string, I can verify that the move constructor is called; however, trying to use the unique_ptr results in the compilation error: error C2248: 'std::unique_ptr<_Ty>::unique_ptr' : cannot access

问题 In C++11 we can transfer the ownership of an object to another unique_ptr using std::move() . After the ownership transfer, the smart pointer that ceded the ownership becomes null and get() returns nullptr. std::unique_ptr<int> p1(new int(42)); std::unique_ptr<int> p2 = std::move(p1); // Transfer ownership What are the situations where this will be useful as it is transferring the ownership to another unique_ptr ? 回答1: The following situations involve transferring ownership from one unique

问题 In C++11 we can transfer the ownership of an object to another unique_ptr using std::move() . After the ownership transfer, the smart pointer that ceded the ownership becomes null and get() returns nullptr. std::unique_ptr<int> p1(new int(42)); std::unique_ptr<int> p2 = std::move(p1); // Transfer ownership What are the situations where this will be useful as it is transferring the ownership to another unique_ptr ? 回答1: The following situations involve transferring ownership from one unique

问题 I have the following factory function: auto factory() -> std::tuple<bool, std::vector<int>> { std::vector<int> vec; vec.push_back(1); vec.push_back(2); return { true, vec }; } auto [b, vec] = factory(); In the return statement is vec considered an xvalue or prvalue and therefore moved or copy elided? My guess is no, because the compiler, when list-initializing the std::tuple in the return statement, still doesn't know that vec is going to be destroyed. So maybe an explicit std::move is

来源： https://stackoverflow.com/questions/64541229/stdmove-in-sub-expressions-of-return-statement

来源： https://stackoverflow.com/questions/64541229/stdmove-in-sub-expressions-of-return-statement

来源： https://stackoverflow.com/questions/64541229/stdmove-in-sub-expressions-of-return-statement

来源： https://stackoverflow.com/questions/63946929/distances-between-embedded-word-vector

问题 This is a general question about symmetry between pointer and reference types in the C++ language. Is this table of correspondence meaningful in C++ (C++11 and beyond)? +-----------+----------------+ | Reference | Pointer | |-----------|----------------| | T& | T* const | | T const& | T const* const | | T&& | ???* | +-----------+----------------+ and if some, what would correspond to ???* ?. (Any additional rows are missing?) ( T&& is for a concrete type T , not a deduced type. See

问题 In C++11, "move semantics" was introduced, implemented via the two special members: move constructor and move assignment. Both of these operations leave the moved-from object constructed. Wouldn't it have been better to leave the source in a destructed state? Isn't the only thing you can do with a moved-from object is destruct it anyway? 回答1: In the "universe of move operations" there are four possibilities: target source is is left ----------------------------------------------------------