copy-elision

C++ return value optimization

生来就可爱ヽ(ⅴ<●) 提交于 2019-11-28 07:17:44
问题 This code: #include <vector> std::vector<float> getstdvec() { std::vector<float> v(4); v[0] = 1; v[1] = 2; v[2] = 3; v[3] = 4; return v; } int main() { std::vector<float> v(4); for (int i = 0; i != 1000; ++i) { v = getstdvec(); } } My incorrect understanding here is that the function getstdvec shouldn't have to actually allocate the vector that it's returning. When I run this in valgrind/callgrind, I see there are 1001 calls to malloc; 1 for the initial vector declaration in main, and 1000

Copy constructor elision? [duplicate]

人走茶凉 提交于 2019-11-27 14:52:32
Possible Duplicate: Why has the destructor been called only once? Given the code below, I fail to understand the output in gcc. I expect two objects to be created and destroyed but instead see only one call to the constructor and the destructor. What's happening here? #include <string> #include <iostream> struct Huge{ Huge() { std::cout << "Constructor" << std::endl; } Huge(Huge const &r) { std::cout << "Copy Constructor" << std::endl; } ~Huge() { std::cout << "Destructor" << std::endl; } }; Huge g() { std::cout << "Entering g" << std::endl; Huge temp; std::cout << "Exiting g" << std::endl;

How to enforce copy elision, why it won't work with deleted copy constructor?

人盡茶涼 提交于 2019-11-27 14:37:43
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()’: t.cc:12:13: error: use of deleted function ‘A::A(const A&)’ A a = fun(); ^ t.cc:4:5: note: declared

Why is RVO disallowed when returning a parameter?

丶灬走出姿态 提交于 2019-11-27 13:11:29
It's stated in [C++11: 12.8/31] : This elision of copy/move operations, called copy elision, is permitted [...] : — in a return statement in a function with a class return type, when the expression is the name of a non-volatile automatic object ( other than a function or catch-clause parameter ) with the same cv-unqualified type as the function return type, the copy/move operation can be omitted by constructing the automatic object directly into the function’s return value This implies #include <iostream> using namespace std; struct X { X() { } X(const X& other) { cout << "X(const X& other)" <

Move Constructor vs Copy Elision. Which one gets called?

爱⌒轻易说出口 提交于 2019-11-27 03:35:26
问题 I have two pieces of code here to show you. They are two classes and each one provides a Move Constructor and a function which returns a temporary. In the first case, the function returning a temporary calls the Move Constructor In the second case, the function returning a temporary just tells the compiler to perform a copy elision I'm confused: in both cases I define a Move Constructor and a random member function returning a temporary. But the behavior changes, and my question is why . Note

What is copy elision and how does it optimize the copy-and-swap idiom?

馋奶兔 提交于 2019-11-26 20:08:28
I was reading Copy and Swap . I tried reading some links on Copy Elision but could not figure out properly what it meant. Can somebody please explain what this optimization is, and especially what is mean by the following text This is not just a matter of convenience but in fact an optimization. If the parameter (s) binds to a lvalue (another non-const object), a copy of the object is made automatically while creating the parameter (s). However, when s binds to a rvalue (temporary object, literal), the copy is typically elided, which saves a call to a copy constructor and a destructor. In the

Copy constructor elision? [duplicate]

半腔热情 提交于 2019-11-26 16:56:45
问题 Possible Duplicate: Why has the destructor been called only once? Given the code below, I fail to understand the output in gcc. I expect two objects to be created and destroyed but instead see only one call to the constructor and the destructor. What's happening here? #include <string> #include <iostream> struct Huge{ Huge() { std::cout << "Constructor" << std::endl; } Huge(Huge const &r) { std::cout << "Copy Constructor" << std::endl; } ~Huge() { std::cout << "Destructor" << std::endl; } };

Why is RVO disallowed when returning a parameter?

蹲街弑〆低调 提交于 2019-11-26 16:18:17
问题 It's stated in [C++11: 12.8/31] : This elision of copy/move operations, called copy elision, is permitted [...] : — in a return statement in a function with a class return type, when the expression is the name of a non-volatile automatic object ( other than a function or catch-clause parameter ) with the same cv-unqualified type as the function return type, the copy/move operation can be omitted by constructing the automatic object directly into the function’s return value This implies

What is copy elision and how does it optimize the copy-and-swap idiom?

情到浓时终转凉″ 提交于 2019-11-26 07:31:37
问题 I was reading Copy and Swap. I tried reading some links on Copy Elision but could not figure out properly what it meant. Can somebody please explain what this optimization is, and especially what is mean by the following text This is not just a matter of convenience but in fact an optimization. If the parameter (s) binds to a lvalue (another non-const object), a copy of the object is made automatically while creating the parameter (s). However, when s binds to a rvalue (temporary object,

How does guaranteed copy elision work?

别说谁变了你拦得住时间么 提交于 2019-11-25 22:27:51
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