[英]c++ recursive function type
羅布·派克提出在2011年(通話鏈接 )約在旅途中,他定義的類型這樣的詞法分析器:
// stateFn represents the state of the scanner
// as a function that returns the next state.
type stateFn func() stateFn
我想在C ++中實現相同的目標,但不知道如何:
// 01: error C3861: 'statefn_t': identifier not found
typedef std::function<statefn_t()> statefn_t;
// 02: error C2371: 'statefn_t': redefinition; different basic types
typedef std::function<class statefn_t()> statefn_t;
// 03: error C2371: 'statefn_t': redefinition; different basic types
typedef std::function<struct statefn_t()> statefn_t;
// 04: error C2065: 'statefn_t': undeclared identifier
typedef std::function<statefn_t*()> statefn_t;
// 05: error C2371: 'statefn_t': redefinition; different basic types
typedef std::function<class statefn_t*()> statefn_t;
// 06: error C2371: 'statefn_t': redefinition; different basic types
typedef std::function<struct statefn_t*()> statefn_t;
注意: 此問題可能已連接(在Rust中是相同的)
編輯:
這是我想要達到的目標:
// statefn_t definition goes here ...
statefn_t* func1()
{
return &func2;
}
statefn_t* func2()
{
return &func1;
}
類型別名不能遞歸。
為了實現一種狀態機,例如go演講中使用的狀態機,您將需要定義一個自定義類型:
class state
{
public:
using fn = std::function<state()>;
state() {}
state(fn f) : f(f){}
operator bool() { return (bool)f; }
operator fn () { return f; }
private:
fn f;
};
用法:
state::fn stateEnd()
{
std::cout << "end\n";
return {};
}
state::fn stateTransit()
{
std::cout << "transit\n";
return stateEnd;
}
state::fn stateStart()
{
std::cout << "start\n";
return stateTransit;
}
int main() {
state::fn s = stateStart;
while(s = s());
}
替代形式:
class state
{
public:
state() {}
template<class T>
state(T&& t) : f(std::forward<T>(t)){}
operator bool() { return (bool)f; }
state operator()() { return f(); }
private:
std::function<state()> f;
};
用法:
state stateEnd()
{
std::cout << "end\n";
return {};
}
state stateTransit()
{
std::cout << "transit\n";
return stateEnd;
}
state stateStart()
{
std::cout << "start\n";
return stateTransit;
}
int main() {
state s {stateStart};
while(s = s());
}
正如Clearer所說,這是一個C ++類型stateFn
的示例,其行為類似於函數,並遞歸返回相同類型的實例。
struct stateFn
{
stateFn& operator() ();
}
如果希望在運行時解析遞歸,同時使其與原始代碼盡可能相似,則可以使用boost :: any或C ++ 17 std :: any,例如:
std::any end(){ std::cout << "end\n"; return {}; }
std::any state(){ std::cout << "some state\n"; return &end; }
std::any begin(){ std::cout << "begin\n"; return &state; }
void advance( std::any& state )
{ state = std::any_cast<std::any(*)()>(state)(); }
int main()
{
for( auto state = begin(); state.has_value(); advance( state ) );
}
如果必須在編譯時解決遞歸,則可以利用自動類型推導:
auto end(){ std::cout << "end\n"; }
auto state(){ std::cout << "some state\n"; return &end; }
auto begin(){ std::cout << "begin\n"; return &state; }
int main()
{
begin()()();
}
當然,這在循環中不起作用,您需要某種編譯時迭代方案才能使其有用...
這不是用C ++做事的自然方法。
在C ++中,它更像是混淆,這體現在難以為狀態推進功能(也就是狀態)尋找好的自描述名稱的過程中:
struct Context {};
class State
{
using F = auto( Context const& ) -> State;
F* next_state_;
public:
auto is_finished() const -> bool { return next_state_ == nullptr; }
auto operator()( Context const& ctx ) const
-> State
{ return next_state_( ctx ); }
State( F* f ): next_state_{ f } {}
inline State();
};
auto intermediate( Context const& ) { return State{ nullptr }; }
auto start( Context const& ) { return State{ intermediate }; }
State::State(): next_state_{ start } {}
#include <iostream>
using namespace std;
auto main() -> int
{
State state;
Context ctx;
cout << boolalpha;
for( ;; )
{
cout << state.is_finished() << endl;
if( state.is_finished() ) { break; }
state = state( ctx );
}
}
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