我有一个Menu<T>
类,其选项是T类型的项目,它可能有Menu<T>
类型的子菜单(对嵌套子菜单的深度没有限制)。
template <typename T>
class Menu {
private:
class Option {
const std::string name;
const T item;
Menu<T>* submenu;
Option* next = nullptr;
friend class Menu<T>;
Option (const std::string& itemName, const T& t, Menu<T>* menu = nullptr) : name(itemName), item(t), submenu(menu) {}
~Option() {if (submenu) delete submenu;}
inline T choose() const;
inline void print (int n) const;
};
Option* first = nullptr; // first option in the menu
Menu<T>* parent = nullptr;
Option* parentOption = nullptr;
enum ChoosingType {Normal, Remove};
public:
Menu() = default;
Menu (const Menu<T>&);
Menu& operator = (const Menu<T>&);
~Menu();
inline void insert (const std::string& itemName, const T& t, Menu<T>* submenu = nullptr, int pos = END_OF_MENU);
T choose() const {return chosenItem().first;}
inline int print() const;
private:
inline std::pair<T, int> chosenItem (ChoosingType = Normal) const;
inline Option* deepCopy (const Option*);
};
我已经测试过它可以正常工作,但我上面的Menu<T>
课程不支持其项目与T不同的子菜单。这个额外的功能非常方便,如果说,主菜单有Action
作为其选项的类型,然后其中一个选项是&#34;取出武器&#34;,其子菜单最好是以Weapon
作为选项,但是就像现在一样,子菜单再次必须有Action
作为选项。
我尝试用
进行推广template <typename T, typename U, typename... Rest>
class Menu { // Menu with T as its option types.
private:
class Option {
const std::string name;
const T item;
Menu<U, Rest...>* submenu; // Submenu with U as its option types.
Option* next = nullptr;
friend class Menu<T, U, Rest...>;
Option (const std::string& itemName, const T& t, Menu<U, Rest...>* menu = nullptr) : name(itemName), item(t), submenu(menu) {}
~Option() {if (submenu) delete submenu;}
inline T choose() const;
inline void print (int n) const;
};
Option* first = nullptr;
// ....
}
int main() {
Menu<int, std::string> menu; // Will not compile.
}
是不正确的,因为Menu<int, std::string> menu;
我尝试使用字符串选项的子菜单创建一个简单的int选项菜单,甚至不会编译,因为子菜单的类型为{{1}这与类模板不匹配。它也没有意义,因为Menu<std::string>
是从它的Menu<int, std::string>
函数返回int,但是进入它的子菜单将返回一个字符串。需要boost :: variant吗?
我只需要有人指出如何开始。我知道这可能看起来属于CodeReview,但是他们只想检查我的代码是否已经有效,但是我在这里尝试概括无法工作(还没有开始),所以我需要这里的专家呼吁如何开始。
更新 根据gmbeard的建议,我使用以下简化代码(真正的菜单类将具有用户将通过输入选择的链接选项列表)。但也有缺点。
choose()
输出:
#include <iostream>
#include <string>
struct Visitor {
virtual void visit (int&) = 0;
virtual void visit (std::string&) = 0;
virtual void visit (char& c) = 0;
};
struct ChooseVisitor : Visitor {
std::pair<int, bool> chosenInt;
std::pair<std::string, bool> chosenString;
std::pair<char, bool> chosenCharacter;
virtual void visit (int& num) override {
chosenInt.first = num;
chosenInt.second = true;
}
virtual void visit (std::string& str) override {
chosenString.first = str;
chosenString.second = true;
}
virtual void visit (char& c) override {
chosenCharacter.first = c;
chosenCharacter.second = true;
}
};
template <typename...> struct Menu;
template <typename T>
struct Menu<T> {
struct Option {
T item;
void accept (ChooseVisitor& visitor) {visitor.visit(item);}
};
Option* option; // Assume only one option for simplicity here.
ChooseVisitor choose() const {
ChooseVisitor visitor;
option->accept(visitor);
return visitor;
}
void insert (const T& t) {option = new Option{t};}
};
// A specialization for the Menu instances that will have submenus.
template <typename T, typename... Rest>
struct Menu<T, Rest...> { // Menu with T as its options type.
struct Option {
T item;
Menu<Rest...>* submenu; // Submenu with the first type in Rest... as its options type.
void accept (ChooseVisitor& visitor) {visitor.visit(item);}
};
Option* option;
ChooseVisitor choose() const {
// In reality there will be user input, of course. The user might not choose to enter a submenu,
// but instead choose from among the options in the current menu.
ChooseVisitor visitor;
if (option->submenu)
return option->submenu->choose();
else
option->accept(visitor);
return visitor;
}
void insert (const T& t, Menu<Rest...>* submenu = nullptr) {option = new Option{t, submenu};}
};
int main() {
Menu<int, std::string, char> menu;
Menu<std::string, char> submenu;
Menu<char> subsubmenu;
subsubmenu.insert('t');
submenu.insert ("", &subsubmenu);
menu.insert (0, &submenu);
const ChooseVisitor visitor = menu.choose();
if (visitor.chosenInt.second)
std::cout << "You chose " << visitor.chosenInt.first << ".\n"; // Do whatever with it.
else if (visitor.chosenString.second)
std::cout << "You chose " << visitor.chosenString.first << ".\n"; // Do whatever with it.
else if (visitor.chosenCharacter.second)
std::cout << "You chose " << visitor.chosenCharacter.first << ".\n"; // Do whatever with it.
}
最大的问题是You chose t.
需要不断更新所有可能的菜单选项类型(最终可能会有数百个数据成员和重载),更不用说可怕的if-checks来获取期望的退货。但是所选项目需要存储,而不仅仅是短期使用。我欢迎改进的想法。
答案 0 :(得分:1)
一种解决方案是创建Menu
的一些部分特化以打开可变参数包。
首先,创建模板类......
// Can be incomplete; only specialized versions will be instantiated...
template<typename... Args> class Menu;
现在,为菜单链的末尾创建一个专门化(没有子菜单)......
template<typename T>
class Menu<T>
{
public:
// No submenu in this specialization
using item_type = T;
std::vector<item_type> items;
...
};
最后,为具有子菜单的Menu
个实例创建一个专门化...
template<typename Head, typename... Tail>
class Menu<Head, Tail...>
{
public:
Menu<Tail...> submenu;
using item_type = Head;
std::vector<item_type> items;
...
};
为简洁起见,这是您的类的简化版本,但如果添加嵌套的Option
类,相同的原则仍然适用。
您可以使用类似的技术通过重载非成员函数来递归子菜单......
template<typename T>
void
print_menu(Menu<T> const& menu)
{
for(auto i : menu.items) {
std::cout << i << std::endl;
}
}
template<typename... T>
void
print_menu(Menu<T...> const& menu)
{
for(auto i : menu.items) {
std::cout << i << std::endl;
}
print_menu(menu.submenu);
}
int
main(int, char*[])
{
Menu<int, std::string> menu{};
menu.items.emplace_back(1);
menu.submenu.items.emplace_back("42");
print_menu(menu);
...
}
更新:choose()
功能的可能实现可以使用访问者模式。您需要为菜单中包含的每种类型(在这种情况下为operator()
和int
)提供重载std::string
的类型...
struct ChooseVisitor
{
void operator()(std::string const& string_val) const
{ /* Do something when a string value is chosen */ }
void operator()(int int_val) const
{ /* Do something when an int value is chosen */ }
};
与print_menu
函数类似,您可以定义几个choose_menu
函数重载...
template<typename... Types, typename Visitor>
void
choose_menu(Menu<Types...> const& menu, Visitor const& visitor)
{
for(auto i : menu.items) {
if(/* is menu item chosen? */) {
visitor(i);
return;
}
}
choose_menu(menu.Submenu, visitor);
}
template<typename Type, typename Visitor>
void
choose_menu(Menu<Type> const& menu, Visitor const& visitor)
{
for(auto i : menu.items) {
if(/* is menu item chosen? */) {
visitor(i);
return;
}
}
}
这样就可以这样使用了......
Menu<int, std::string> menu{};
...
choose_menu(menu, ChooseVisitor{});
要为choose()
函数推导出您的想法有点困难,但您应该能够根据大多数情况调整上述内容。
答案 1 :(得分:1)
我有一种轻微的感觉,这里的事情已被过度设计。
正如您所推测的,这是一个使用Boost Variant的版本。
我简化了一些事情。特别是我倾向于初始化列表构造函数,因此您可以简单地以递归方式构造菜单树:
Menu menu = MenuT<int> {
{ "one", 1 },
{ "two", 2 },
{ "three", 3 },
{ "forty-two", 42,
Menu(MenuT<std::string> {
{"Life, The Universe And Everything", "LtUae"},
{"Dent", "Arthur",
Menu(MenuT<bool> {
{"yes", true},
{"no", false},
})
},
})
},
};
如您所见,它会在不同级别混合MenuT<int>
,MenuT<std::string>
和MenuT<bool>
。你可以毫不费力地访问它:
struct simple : boost::static_visitor<>
{
void operator()(Menu& m) const { boost::apply_visitor(*this, m); }
template <typename T> void operator()(MenuT<T>& m) const {
std::cout << "-----------\n";
for (auto& o : m.options) {
std::cout << "option '" << o.name << "':\t" << o.item << "\n";
if (o.submenu)
(*this)(*o.submenu);
}
}
};
打印
-----------
option 'one': 1
option 'two': 2
option 'three': 3
option 'forty-two': 42
-----------
option 'Life, The Universe And Everything': LtUae
option 'Dent': Arthur
-----------
option 'yes': true
option 'no': false
<强> Live On Coliru 强>
#include <string>
#include <vector>
#include <boost/optional.hpp>
#include <boost/variant.hpp>
#include <iostream>
template <typename> struct MenuT;
using Menu = boost::make_recursive_variant <
boost::recursive_wrapper<MenuT<int>>,
boost::recursive_wrapper<MenuT<std::string>>,
boost::recursive_wrapper<MenuT<bool>>
>::type;
template <typename T> struct MenuT {
struct Option {
std::string name;
T item;
boost::optional<Menu> submenu;
Option(std::string name, T t, boost::optional<Menu> submenu = boost::none)
: name(name), item(t), submenu(submenu)
{ }
T choose() const;
void print(int n) const;
};
private:
template <typename U> friend struct MenuT;
friend struct visitors;
std::vector<Option> options;
//boost::optional<Menu&> parent; // TODO e.g. link_parents_visitor
enum ChoosingType { Normal, Remove };
public:
enum SpecialPosition : size_t { END_OF_MENU = size_t(-1) };
MenuT() = default;
MenuT(std::initializer_list<Option> options) : options(options) {}
void insert(const std::string &itemName, const T &t, boost::optional<Menu> submenu = boost::none, size_t pos = END_OF_MENU) {
auto it = (pos == END_OF_MENU
? options.end()
: std::next(options.begin(), std::min(pos, options.size())));
options.emplace(it, itemName, t, submenu);
}
T choose() const { return chosenItem().first; }
int print() const;
private:
std::pair<T, int> chosenItem(ChoosingType = Normal) const;
};
struct visitors {
struct simple : boost::static_visitor<>
{
void operator()(Menu& m) const { boost::apply_visitor(*this, m); }
template <typename T> void operator()(MenuT<T>& m) const {
std::cout << "-----------\n";
for (auto& o : m.options) {
std::cout << "option '" << o.name << "':\t" << o.item << "\n";
if (o.submenu)
(*this)(*o.submenu);
}
}
};
};
static const visitors::simple demo { };
int main()
{
Menu menu = MenuT<int> {
{ "one", 1 },
{ "two", 2 },
{ "three", 3 },
{ "forty-two", 42,
Menu(MenuT<std::string> {
{"Life, The Universe And Everything", "LtUae"},
{"Dent", "Arthur",
Menu(MenuT<bool> {
{"yes", true},
{"no", false},
})
},
})
},
};
std::cout << std::boolalpha;
demo(menu);
}
<强>说明强>
值得注意的遗漏是我没有初始化
boost::optional<Menu&> parent; // TODO e.g. link_parents_visitor
我认为你仍然想要简化这一点(我想你可能不希望在一开始就将MenuT和Option分开。你刚才介绍了它试图嵌套不同菜单选项的类型?