c++14

问题 Discussion According to the standard §20.10.2/1 Header <type_traits> synopsis [meta.type.synop]: 1 The behavior of a program that adds specializations for any of the class templates defined in this subclause is undefined unless otherwise specified. This specific clause contradicts to the general notion that STL should be expandible and prevents us from expanding type traits as in the example below: namespace std { template< class T > struct is_floating_point<std::complex<T>> : std::integral

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 3 years ago . I read a c++ tutorial that says arithmetic operators return the smallest data type possible (i.e. if 2 ints are added the return type will be int, if a float and a double are added the return type will be double). However, it also said that arithmetic operations on shorts return ints. Considering

问题 Basically in below I want to see if I can get around having to use auto keyword Suppose that we have the following piece of code [works with g++ 4.9.2 (Ubuntu 4.9.2-10ubuntu13) & clang version 3.6.0] : //g++ -std=c++14 test.cpp //test.cpp #include <iostream> using namespace std; template<typename T> constexpr auto create() { class test { public: int i; virtual int get(){ return 123; } } r; return r; } auto v = create<int>(); int main(void){ cout<<v.get()<<endl; } How can I specify the type of

问题 Consider this working code: #include <iostream> #include <utility> #include <array> template <typename... Args> void foo (Args&&... args) { const auto v = {args...}; for (auto x : v) std::cout << x << ' '; std::cout << '\n'; } template <typename> struct Foo; template <std::size_t... Is> struct Foo<std::index_sequence<Is...>> { template <typename Container> static void execute (const Container& v) { foo(v[Is]...); } }; template <std::size_t N> void fooArray (const std::array<int, N>& a) { Foo

问题 enum Symbols { BAR, BELL, PLUM, ORANGE, CHERRY, DOESNOTMATTER, }wheel1, wheel2, wheel3; map<Symbols[3], int> symb = {{{BAR, BAR, BAR}, 250}, {{BELL, BELL, BELL}, 20}, {{BELL, BELL, BAR}, 20}, {{PLUM, PLUM, BAR}, 14}, {{PLUM, PLUM, PLUM}, 14}, {{ORANGE, ORANGE, BAR}, 10}, {{ORANGE, ORANGE, ORANGE}, 10}, {{CHERRY, CHERRY, CHERRY}, 7}, {{CHERRY, CHERRY, DOESNOTMATTER}, 5}, {{CHERRY, DOESNOTMATTER, DOESNOTMATTER}, 2}}; I have defined an enumerator named Symbols. I am trying to create a map, who's

问题 I tried to force explicit conversion in return statement by means of using of uniform initialization syntax as it is in following: #include <iostream> #include <cstdlib> struct A { explicit operator bool () const { return false; } }; bool f() { return {A{}}; // error: no viable conversion from 'void' to 'bool' // equivalent to return bool{A{}}; at my mind } bool g() { return static_cast< bool >(A{}); } struct B { B(A) { ; } }; B h() { return {A{}}; // equivalent to return B{A{}}; } inline std

问题 Can someone help me to understand why the following code causes a warning struct A { A() : _a( 0 ) {} const int& _a; }; int main() { A a; } with warning warning: binding reference member '_a' to a temporary value [-Wdangling-field] A() : _a( 0 ) {} but this code, where std::move is used to initialize the member _a , does not: struct A { A() : _a( std::move(0) ) {} const int& _a; }; int main() { A a; } Aren't 0 and std::move( 0 ) both r-values? 回答1: This is an expression: 0 It's a very small

问题 I am struggeling writing a identifier parser, which parses a alphanum string which is not a keyword. the keywords are all in a table: struct keywords_t : x3::symbols<x3::unused_type> { keywords_t() { add("for", x3::unused) ("in", x3::unused) ("while", x3::unused); } } const keywords; and the parser for a identifier should be this: auto const identifier_def = x3::lexeme[ (x3::alpha | '_') >> *(x3::alnum | '_') ]; now i try to combine these so an identifier parser fails on parsing a keyword. I

问题 How to determine whether a pointer of base ( B ) class is (polymorphism-ly) override a certain virtual function of the base class? class B{ public: int aField=0; virtual void f(){}; }; class C : public B{ public: virtual void f(){aField=5;}; }; class D: public B{}; int main() { B b; C c; D d; std::vector<B*> bs; bs.push_back(&b); bs.push_back(&c); bs.push_back(&d); for(int n=0;n<3;n++){ //std::cout<<(bs[n]->f)==B::f<<std::endl; //should print "true false true" } } I tried to compare the

问题 Let's say we want to create a helper class to reverse template pack e.g. as follows: #include <tuple> #include <utility> #include <typeinfo> #include <iostream> template <class> struct sizer; template <template<class...> class Pack, class... Args> struct sizer<Pack<Args...>> { static constexpr size_t value = sizeof...(Args); }; template <class Pack, class Indices = std::make_index_sequence<sizer<Pack>::value>> struct reverse_pack; template <class... Args, size_t... I> struct reverse_pack<std: