Advantages of using std::make_unique over new operator The difference is that std::make_unique returns an object of type std::unique_ptr and new returns a pointer to the created object For memory allocation failures, they will both throw
Why use std::make_unique in C++17? - Stack Overflow Since then, C++17 has clarified the evaluation order, making Syntax A safe too, so here's my question: is there still a reason to use std::make_unique over std::unique_ptr 's constructor in C++17? Can you give some examples? As of now, the only reason I can imagine is that it allows to type MyClass only once (assuming you don't need to rely on polymorphism with std::unique_ptr<Base>(new
Differences between std::make_unique and std::unique_ptr with new How would you deal with polymorphism without new? For example, you need a container of entities, the exact type of which is defined at runtime, and use std::unique_ptr rather than raw pointers: std::vector< std::unique_ptr<Base> > entities; You have to use entities emplace_back(new Derived); if you want to add elements of the derived class, as far as I can think
How can I use make_unique with c++11? - Stack Overflow If you look at the cppreference doc for std::make_unique(), it shows a possible implementation that (with minor tweaks) can be applied to C++11 code If your code doesn't need to worry about std::unique<T[]> support for arrays, then the simplest implementation would look like this:
c++ - make_uniqueの利点 - スタック・オーバーフロー make_unique を使用するとコンストラクター呼び出しではなくなるので auto 変数が使えます。対して std::unique_ptr はコンストラクター呼び出しなので型名を省略できません。 std::unique_ptr<int> pInt {new int{1}}; auto pInt = std::make_unique<int>(1); また前者は std::unique_ptr でポインター管理され自動的に delete さ
Custom initialize array with std::make_unique - Stack Overflow Say I would like to create a std::unique_ptr<int[]>, but I would like to initialize the created array to custom values: {1,2,3,4,5} I can use new and pass the raw pointer to std::unique_ptr constructor which will then own and manage it
c++ - Does assigning make_unique require std::move () to an empty . . . It's not, because in both cases the expression may bind to an rvalue reference In the first, std::make_unique already returns a pure rvalue In the second, std::move does a cast to an rvalue reference, it's redundant It also doesn't matter if the destination object is empty or not The mere act of invoking assignment doesn't depend on how the reference it received was bound The result would