copy-elision

问题 At the 2016 Oulu ISO C++ Standards meeting, a proposal called Guaranteed copy elision through simplified value categories was voted into C++17 by the standards committee. How exactly does guaranteed copy elision work? Does it cover some cases where copy elision was already permitted, or are code changes needed to guarantee copy elision? 回答1: Copy elision was permitted to happen under a number of circumstances. However, even if it was permitted, the code still had to be able to work as if the

问题 I see a lot of code at work where people use emplace and emplace_back with a temporary object, like this: struct A { A::A(int, int); }; vector<A> v; vector<A>.emplace_back(A(1, 2)); I know that the whole point of emplace_back is to be able to pass the parameters directly, like this: v.emplace_back(1, 2); But unfortunately this is not clear to a few people. But let's not dwell on that.... My question is: is the compiler able to optimize this and skip the create and copy? Or should I really try

问题 I see a lot of code at work where people use emplace and emplace_back with a temporary object, like this: struct A { A::A(int, int); }; vector<A> v; vector<A>.emplace_back(A(1, 2)); I know that the whole point of emplace_back is to be able to pass the parameters directly, like this: v.emplace_back(1, 2); But unfortunately this is not clear to a few people. But let's not dwell on that.... My question is: is the compiler able to optimize this and skip the create and copy? Or should I really try

问题 I see a lot of code at work where people use emplace and emplace_back with a temporary object, like this: struct A { A::A(int, int); }; vector<A> v; vector<A>.emplace_back(A(1, 2)); I know that the whole point of emplace_back is to be able to pass the parameters directly, like this: v.emplace_back(1, 2); But unfortunately this is not clear to a few people. But let's not dwell on that.... My question is: is the compiler able to optimize this and skip the create and copy? Or should I really try

问题 I see a lot of code at work where people use emplace and emplace_back with a temporary object, like this: struct A { A::A(int, int); }; vector<A> v; vector<A>.emplace_back(A(1, 2)); I know that the whole point of emplace_back is to be able to pass the parameters directly, like this: v.emplace_back(1, 2); But unfortunately this is not clear to a few people. But let's not dwell on that.... My question is: is the compiler able to optimize this and skip the create and copy? Or should I really try

问题 I have an uncopiable class. Copying this would be problematic. I want to guarantee that it won't be ever copied, so I made its copy constructor deleted : class A { public: A(); A(const A&) = delete; }; A fun() { return A(); }; int main() { A a = fun(); }; Unfortunately, g++ won't compile this on the reason: t.cc: In function ‘A fun()’: t.cc:8:12: error: use of deleted function ‘A::A(const A&)’ return A(); ^ t.cc:4:5: note: declared here A(const A&) = delete; ^ t.cc: In function ‘int main()’:

问题 This question already has answers here : How to initialize array of classes with deleted copy constructor (C++11) (2 answers) Closed 25 days ago . As far as I understand the new rules correctly https://en.cppreference.com/w/cpp/language/copy_elision http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/p0135r0.html This code should compile for C++17 standard conform compilers struct NonTrivialClass { ~NonTrivialClass( ){ } }; class MainNonTrivialClass { public: MainNonTrivialClass(int t) :

问题 There is a paragraph about guaranteed copy elision in c++ draft n4606 [dcl.init] 17.6: If the destination type is a (possibly cv-qualified) class type: If the initializer expression is a prvalue and the cv-unqualified version of the source type is the same class as the class of the destination, the initializer expression is used to initialize the destination object. [ Example : T x = T(T(T())); calls the T default constructor to initialize x . — end example ] [...] There is also a Q&A talks

问题 Copy elision is a neat optimization technique and in some cases relying on copy elision can actually be faster than passing around references "by hand". So, let's assume you have identified a critical code path where you rely on the fact that the copy elision is performed by your compiler for the code path for maximum performance. But now you are relying on a compiler optimization. Is there any (compiler specific, obviously) way to ensure that the copy elision is actually performed and have

问题 I have an uncopiable class. Copying this would be problematic. I want to guarantee that it won't be ever copied, so I made its copy constructor deleted : class A { public: A(); A(const A&) = delete; }; A fun() { return A(); }; int main() { A a = fun(); }; Unfortunately, g++ won't compile this on the reason: t.cc: In function ‘A fun()’: t.cc:8:12: error: use of deleted function ‘A::A(const A&)’ return A(); ^ t.cc:4:5: note: declared here A(const A&) = delete; ^ t.cc: In function ‘int main()’: