c++14

问题 Say I have a class representing automata, whose states are numbered ( using state_t = unsigned ) and whose transitons are also numbered ( using transition_t = unsigned ). Of course at some point I end up messing some calls because transition_t and state_t are the same type, so the compiler does not enforce the (semantic) type safety. That's easy to workaround by using a small class templated by a tag ( struct transition_tag {}; struct state_tag {}; ), so now transition_t and state_t are

问题 I frequently come across a situation where we create a class that acts on some enumeration, but later we derive and we want to add more values to the enumeration without changing the base class. I see this question from 2009: Base enum class inheritance However, I know there were a number of changes to enum in C++11, 14, 17. Do any of those changes allow for extension of enums from base class to derived? class Base { enum State {STATE_1, STATE_2, STATE_3}; }; class Derived : public Base {

问题 I frequently come across a situation where we create a class that acts on some enumeration, but later we derive and we want to add more values to the enumeration without changing the base class. I see this question from 2009: Base enum class inheritance However, I know there were a number of changes to enum in C++11, 14, 17. Do any of those changes allow for extension of enums from base class to derived? class Base { enum State {STATE_1, STATE_2, STATE_3}; }; class Derived : public Base {

问题 I frequently come across a situation where we create a class that acts on some enumeration, but later we derive and we want to add more values to the enumeration without changing the base class. I see this question from 2009: Base enum class inheritance However, I know there were a number of changes to enum in C++11, 14, 17. Do any of those changes allow for extension of enums from base class to derived? class Base { enum State {STATE_1, STATE_2, STATE_3}; }; class Derived : public Base {

问题 I frequently come across a situation where we create a class that acts on some enumeration, but later we derive and we want to add more values to the enumeration without changing the base class. I see this question from 2009: Base enum class inheritance However, I know there were a number of changes to enum in C++11, 14, 17. Do any of those changes allow for extension of enums from base class to derived? class Base { enum State {STATE_1, STATE_2, STATE_3}; }; class Derived : public Base {

问题 In the simple parser library I am writing, the results of multiple parsers is combined using std::tuple_cat . But when applying a parser that returns the same result multiple times, it becomes important to transform this tuple into a container like a vector or a deque. How can this be done? How can any tuple of the kind std::tuple<A> , std::tuple<A, A> , std::tuple<A, A, A> etc be converted into a std::vector<A> ? I think this might be possible using typename ...As and sizeof ...(As) , but I

问题 In the simple parser library I am writing, the results of multiple parsers is combined using std::tuple_cat . But when applying a parser that returns the same result multiple times, it becomes important to transform this tuple into a container like a vector or a deque. How can this be done? How can any tuple of the kind std::tuple<A> , std::tuple<A, A> , std::tuple<A, A, A> etc be converted into a std::vector<A> ? I think this might be possible using typename ...As and sizeof ...(As) , but I

问题 In the simple parser library I am writing, the results of multiple parsers is combined using std::tuple_cat . But when applying a parser that returns the same result multiple times, it becomes important to transform this tuple into a container like a vector or a deque. How can this be done? How can any tuple of the kind std::tuple<A> , std::tuple<A, A> , std::tuple<A, A, A> etc be converted into a std::vector<A> ? I think this might be possible using typename ...As and sizeof ...(As) , but I

问题 The short version of my question is this: How can I use something like std::bind() with a standard library algorithm? Since the short version is a bit devoid of details, here is a bit of an explanation: Assume I have the algorithms std::transform() and now I want to implement std::copy() (yes, I realize that there is std::copy() in the standard C++ library). Since I'm hideously lazy, I clearly want to use the existing implementation of std::transform() . I could, of course, do this: struct

问题 The short version of my question is this: How can I use something like std::bind() with a standard library algorithm? Since the short version is a bit devoid of details, here is a bit of an explanation: Assume I have the algorithms std::transform() and now I want to implement std::copy() (yes, I realize that there is std::copy() in the standard C++ library). Since I'm hideously lazy, I clearly want to use the existing implementation of std::transform() . I could, of course, do this: struct