copy-elision

问题 I push_back a temporary object into a vector like this, vector<A> vec; vec.push_back(A("abc")); will the compiler apply copy-elision to construct the temporary A("abc") directly into the vector , so that A 's copy ctor won't be triggered when pushing the temporary object into vec . 回答1: If you have a compiler that supports rvalue references, it will be moved into the vector, which is sometimes quite cheap. An alternative to that is to directly construct the object in the vector, which can be

问题 Consider the following: struct X { X() {} X(X&&) { puts("move"); } }; X x = X(); In C++14, the move could be elided despite the fact that the move constructor has side effects thanks to [class.copy]/31, This elision of copy/move operations ... is permitted in the following circumstances ... when a temporary class object that has not been bound to a reference (12.2) would be copied/moved to a class object with the same cv-unqualified type In C++17 this bullet was removed. Instead the move is

问题 I have come across the situation where I really do need to execute non-trivial code in a copy-constructor/assignment-operator. The correctness of the algorithm depends on it. While I could disable return value optimisation with a compiler switch, it seems a waste because it's only the one type I need it disabled for, so why should the performance of the whole application suffer? (Not to mention that my company would not allow me to add the switch, anyway). struct A { explicit A(double val) :

问题 Consider the following program: #include<iostream> using namespace std; struct S { S() = default; S(const S& other) = delete; S(S&& other) = delete; int i; }; S nakedBrace() { return {}; // no S constructed here? } S typedBrace() { return S{}; } int main() { // produce an observable effect. cout << nakedBrace().i << endl; // ok cout << typedBrace().i << endl; // error: deleted move ctor } Sample session: $ g++ -Wall -std=c++14 -o no-copy-ctor no-copy-ctor.cpp no-copy-ctor.cpp: In function 'S

问题 What is copy elision? What is (named) return value optimization? What do they imply? In what situations can they occur? What are limitations? If you were referenced to this question, you're probably looking for the introduction. For a technical overview, see the standard reference. See common cases here. 回答1: Introduction For a technical overview - skip to this answer. For common cases where copy elision occurs - skip to this answer. Copy elision is an optimization implemented by most

问题 For example: In accepted answer https://stackoverflow.com/a/14623480/1423254, Does copy elision and RVO would still work for classes without move constructors? Yes, RVO still kicks in. Actually, the compiler is expected to pick: RVO (if possible) In accepted answer https://stackoverflow.com/a/38043447/1423254, Under non-guaranteed copy elision rules, this will create a temporary, then move from that temporary into the function's return value. That move operation may be elided, but T must

问题 What is copy elision? What is (named) return value optimization? What do they imply? In what situations can they occur? What are limitations? If you were referenced to this question, you're probably looking for the introduction. For a technical overview, see the standard reference. See common cases here. 回答1: Introduction For a technical overview - skip to this answer. For common cases where copy elision occurs - skip to this answer. Copy elision is an optimization implemented by most

问题 I want to know when copy/move elision applies (or is allowed to apply) to explicitly delete d copy/move constructors and to non- delete d copy/move constructors. Here are the specifics: Can an explicitly delete d copy ctor or move ctor get elided? Is an attempt to construct an object from another same-type object or temporary object ever allowed to succeed by skipping over the delete d copy ctor and/or delete d move ctor? Here’s what happens in VC12 (with which I’m not sure if there is an

问题 Suppose I have this hypothetical, odd and unintuitive situation #include <iostream> struct A { A() { member = 1; } A(const A &) { member = 2; } int member; }; int main() { A a = A(); A b = a; std::cout << a.member << std::endl; std::cout << b.member << std::endl; return 0; } I know that copy elision means that a will be initialized with just the default constructor and that b will be initialized with the copy constructor. I also know that (on gcc at least) you can tell the compiler not to do

问题 My question is different because I may "know" copy-elision. I am learning copy initialization. However,the following code confused me because I have already turned off the copy-elision using -fno-elide-contructors -O0 option. #include <iostream> using namespace std; class test{ public : test(int a_, int b_) : a{a_}, b{b_} {} test(const test& other) { cout << "copy constructor" << endl; } test& operator=(const test& other) { cout << "copy assignment" << endl; return *this; } test(test&& other)