move

问题 I read in another question that when implementing a move constructor it is good practice to std::move each member in the initializer list because if the member happens to be another object then that objects move constructor will be called. Like so... //Move constructor Car::Car(Car && obj) : prBufferLength(std::move(obj.prBufferLength)), prBuffer(std::move(obj.prBuffer)) { obj.prBuffer = nullptr; obj.prBufferLength = 0; } However in all the sample move assignment operators I've seen, there

问题 I have a silly example here, just to demonstrate an issue I'm running into with another library and pattern matching. struct Person { name: String, age: i32, choice: Choices } #[derive(Debug)] enum Choices { Good, Neutral, Evil } fn find(p: Person) { match (p.choice, p.age) { (Choices::Good, a) if a < 80 => { announce(p); } (_, a) if a >= 80 => { println!("You're too old to care."); } _ => { println!("You're not very nice!") } } } fn announce(p: Person) { println!("Your name is {}. You are {:

问题 This question already has answers here : What are copy elision and return value optimization? (4 answers) Copy constructor not called? (2 answers) Closed 3 years ago . As per my understanding, the move constructor will be called when there is a temporary object created. Here the getA() function is returning a temporary object but my program is not printing the message from the move constructor: #include <iostream> using namespace std; class A { public: A() { cout<<"Hi from default\n"; } A(A &

问题 If I have an object e of type Error which implements a move constructor, will throwing std::move( e ) use the move constructor of Error to "duplicate" e , so does it avoid making an actual copy of the object? So if I have Error e; throw std::move( e ); will the copy constructor of Error be called or not? This is of interest when your move constructor is noexcept (as it should be), but your copy constructor isn't. 回答1: § 15.1 [except.throw]: Throwing an exception copy-initializes (8.5, 12.8) a

问题 I read this (incredibly well written) article about Forwarding Reference in C++11 by Scott Meyers. Now, focus on this part of the article: template <class... Args> void emplace_back(Args&&... args); // deduced parameter types ⇒ type deduction; ... // && ≡ universal references So, in contrast with other cases, the ellipses doesn't make the && an rvalue reference, but it's still universal references. From what I've understood, when we have universal references, we can call the function passing

问题 All my files are in specific folders: 17\1\1\PRO 17\1\2\PRO 17\2\1\PRO xx\xx\xx\PRO 17 is the year (so 18 for next year etc) the first 1 is the folder specifying the case number (can be up to 100). The second 1 is the sub parts on the case number. That last 1 has a folder PRO in it where all data resides. We need to move these files, but the files need to stay inside their respective "PRO" folders. For example: a file in 17\1\1\pro\xxx\www\ needs to go to 17\1\1\pro\movies a file in 17\2\2

问题 Hi i'm working on a simple program and for the set up of the program i need the program to check a directory for zip files and any zip files in there need to be moved into another folder. Lets say i have folder1 and it contains 6 zip files and then i have another folder called folder2 that i need all the zips and only the zips in folder1 moved to folder2 Thank you for any help one this problem. Btw i'm a noob so any code samples would be greatly appreciated 回答1: For each file in folder1 , use

问题 Here is my test code: void test(std::vector<int> vec){}; void test(std::vector<int> && vec){}; int main(int argc, char * argv[]) { std::vector<int> v; test(v); test(std::move(v)); return 0; } When I try to call test(std::move(v)) , I was told test is multiply implemented. Obviously I have used std::move making v a rvalue. Won't test(std::vector<int> &&) be called specifically? 回答1: This isn't directly related to rvalues, or moving. The same happens with an lvalue reference overload void test

问题 Here is my test code: void test(std::vector<int> vec){}; void test(std::vector<int> && vec){}; int main(int argc, char * argv[]) { std::vector<int> v; test(v); test(std::move(v)); return 0; } When I try to call test(std::move(v)) , I was told test is multiply implemented. Obviously I have used std::move making v a rvalue. Won't test(std::vector<int> &&) be called specifically? 回答1: This isn't directly related to rvalues, or moving. The same happens with an lvalue reference overload void test

问题 Here is a test file from gcc, live demo struct do_nothing { template <class T> void operator()(T*) {} }; int main() { int i = 0; std::unique_ptr<int, do_nothing> p1(&i); std::unique_ptr<int> p2; static_assert(!std::is_assignable<decltype(p2), decltype(p1)>::value, ""); // note ! here. } std::is_assignable If the expression std::declval<T>() = std::declval<U>() is well-formed in unevaluated context, provides the member constant value equal true. Otherwise, value is false. Access checks are