c++14

问题 I am confused about using std::move() in below code: If I uncomment line at (2) the output would be: 1 2 3 but if I uncomment line at (1) output would be nothing which means that move constructor of std::vector was called! Why do we have to make another call to std::move at (1) to make move constructor of std::vector to be called? What I understood that std::move get the r-value of its parameter so, why we have to get the r-value of r-value at (1)? I think this line _v = rv; at (2) is more

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 The directive: #ifndef __cplusplus #error C++ is required #elif __cplusplus < 201402L #error C++14 is required #endif The command-line: g++ -Wall -Wextra -std=c++14 -c -o header.o header.hpp My g++ version: g++ (tdm-1) 4.9.2 The error C++14 is required is generated even when I added -std=c++14 , I don't know why. Please tell me how to fix this. 回答1: According to the GCC CPP manual (version 4.9.2 and 5.1.0): __cplusplus This macro is defined when the C++ compiler is in use. You can use _

问题 The directive: #ifndef __cplusplus #error C++ is required #elif __cplusplus < 201402L #error C++14 is required #endif The command-line: g++ -Wall -Wextra -std=c++14 -c -o header.o header.hpp My g++ version: g++ (tdm-1) 4.9.2 The error C++14 is required is generated even when I added -std=c++14 , I don't know why. Please tell me how to fix this. 回答1: According to the GCC CPP manual (version 4.9.2 and 5.1.0): __cplusplus This macro is defined when the C++ compiler is in use. You can use _

问题 The directive: #ifndef __cplusplus #error C++ is required #elif __cplusplus < 201402L #error C++14 is required #endif The command-line: g++ -Wall -Wextra -std=c++14 -c -o header.o header.hpp My g++ version: g++ (tdm-1) 4.9.2 The error C++14 is required is generated even when I added -std=c++14 , I don't know why. Please tell me how to fix this. 回答1: According to the GCC CPP manual (version 4.9.2 and 5.1.0): __cplusplus This macro is defined when the C++ compiler is in use. You can use _

问题 There's two containers: owner and non-owner of resource. As I have only 1 owner, I suppose I need unique_ptr. class OwnershipContainer { public: void add(std::unique_ptr<Obj> obj) { objects.push_back(std::move(obj)); } Obj* get(std::size_t i) { return objects[i].get(); } private: std::vector<std::unique_ptr<Obj>> objects; }; What of kind of pointer i have to use for non-owner container? First thought was raw pointer. But i cannot give guarantee that lifetime of Obj match or exceed lifetime of

问题 The following code compiles fine in GCC (4.9.3) and VC++ (19.00.23506) but gives these error in Clang (3.7.0). error: constexpr function's return type 'Foo' is not a literal type note: 'Foo' is not literal because it is not an aggregate and has no constexpr constructors other than copy or move constructors Code: #include <iostream> #include <vector> struct Foo { std::vector<int> m_vec; Foo(const int *foo, std::size_t size=0):m_vec(foo, foo+size) {;} //Foo(const std::initializer_list<int>