迭代不同的类型

Question

给出以下代码:

struct Window{
    void show();
    //stuff
}w1, w2, w3;

struct Widget{
    void show();
    //stuff
}w4, w5, w6;

struct Toolbar{
    void show();
    //stuff
}t1, t2, t3;

我想要show一堆物品:

for (auto &obj : {w3, w4, w5, t1})
    obj.show();

然而,这不会编译,因为std::initializer_list<T>在for-loop中无法推断T,事实上并没有真正T适合的.我不想创建类型擦除类型,因为需要大量的代码和不必要的运行时开销.如何正确编写循环以便obj分别为概念列表中的每个项推导出类型？

Answer 1

在现代C++中,您将使用折叠表达式来"遍历"应用成员函数的异构参数:

auto Printer = [](auto&&... args) {
    (args.show(), ...);
};

Printer(w1, w2, w3, w4, w5, w6, t1, t2, t3);

演示

您可以在我的博客中阅读更多相关内容

Answer 2

boost :: fusion很棒但是oldskool - 它迎合了c ++ 03中的不足.

c ++ 11的可变参数模板扩展到救援!

#include <iostream>

struct Window{
    void show() {
        std::cout << "Window\n";
    }
    //stuff
}w1, w2, w3;

struct Widget{
    void show() {
        std::cout << "Widget\n";
    }
    //stuff
}w4, w5, w6;

struct Toolbar{
    void show()
    {
        std::cout << "Toolbar\n";
    }
    //stuff
}t1, t2, t3;


template<class...Objects>
void call_show(Objects&&...objects)
{
    using expand = int[];
    (void) expand { 0, ((void)objects.show(), 0)... };
}

auto main() -> int
{
    call_show(w3, w4, w5, t1);
    return 0;
}

预期产量:

Window
Widget
Widget
Toolbar

另一种更通用的方式(需要c ++ 14):

// note that i have avoided a function names that look like
// one in the standard library.

template<class Functor, class...Objects>
void for_all(Functor&& f, Objects&&... objects)
{
    using expand = int[];
    (void) expand { 0, (f(std::forward<Objects>(objects)), 0)... };

}

如此称呼:

for_all([](auto& thing) { thing.show(); }, w3, w4, w5, t1);

Answer 3

另一种选择是使用boost::tupleor std::tuple和boost::fusion::for_each算法:

#include <boost/fusion/algorithm/iteration/for_each.hpp>
#include <boost/fusion/adapted/boost_tuple.hpp>

boost::fusion::for_each(
    boost::tie(w1, w2, w3, w4, w5, w6, t1, t2, t3), // by reference, not a copy
    [](auto&& t) { t.show(); } 
    );

出于好奇,将Richard Hodges的方法生成的装配输出与上述方法进行了比较.与gcc-4.9.2 -Wall -Wextra -std=gnu++14 -O3 -march=native生产的汇编代码相同.

Answer 4

基于/sf/answers/482610551/,这可以在不创建额外功能,增强或继承的情况下工作.

标题:

#include <tuple>
#include <utility> 

template<std::size_t I = 0, typename FuncT, typename... Tp>
inline typename std::enable_if<I == sizeof...(Tp), void>::type
  for_each(const std::tuple<Tp...> &, FuncT) // Unused arguments are given no names.
  { }

template<std::size_t I = 0, typename FuncT, typename... Tp>
inline typename std::enable_if<I < sizeof...(Tp), void>::type
  for_each(const std::tuple<Tp...>& t, FuncT f)
  {
    f(std::get<I>(t));
    for_each<I + 1, FuncT, Tp...>(t, f);
  }

template<std::size_t I = 0, typename FuncT, typename... Tp>
inline typename std::enable_if<I == sizeof...(Tp), void>::type
  for_each(std::tuple<Tp...> &&, FuncT) // Unused arguments are given no names.
  { }

template<std::size_t I = 0, typename FuncT, typename... Tp>
inline typename std::enable_if<I < sizeof...(Tp), void>::type
  for_each(std::tuple<Tp...>&& t, FuncT f)
  {
    f(std::get<I>(t));
    for_each<I + 1, FuncT, Tp...>(std::move(t), f);
  }

的.cpp:

struct Window{
    void show(){}
    //stuff
}w1, w2, w3;

struct Widget{
    void show(){}
    //stuff
}w4, w5, w6;

struct Toolbar{
    void show(){}
    //stuff
}t1, t2, t3;

int main() {
    for_each(std::tie(w3, w4, w5, t1), [](auto &obj){
        obj.show();
    });
}

Answer 5

Window,Widget并Toolbar共享通用接口,因此您可以创建抽象类并使其他类继承它:

struct Showable {
    virtual void show() = 0; // abstract method
};

struct Window: Showable{
    void show();
    //stuff
}w1, w2, w3;

struct Widget: Showable{
    void show();
    //stuff
}w4, w5, w6;

struct Toolbar: Showable{
    void show();
    //stuff
}t1, t2, t3;

然后,您可以创建指针数组Showable并迭代它:

int main() {
    Showable *items[] = {&w3, &w4, &w5, &t1};
    for (auto &obj : items)
        obj->show();
}

看到它在线工作

Answer 6

我推荐Boost.Hana,其中恕我直言是最好,最灵活的模板元编程库.

#include <boost/hana/ext/std/tuple.hpp>
#include <boost/hana.hpp>

namespace hana = boost::hana;

hana::for_each(std::tie(w3, w4, w5, t1), [](auto& obj) { obj.show(); });