c ++递归函数类型

Question

Rob Pike在2011年（link）就go中的词法分析者发表了演讲，在那里他定义了这样一种类型：

// stateFn represents the state of the scanner
// as a function that returns the next state.
type stateFn func() stateFn

我想在C ++中实现相同，但无法弄清楚如何：

// 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：this问题可能已连接（在Rust中也是如此）

编辑：

这是我想要实现的目标：

// statefn_t definition goes here ...

statefn_t* func1()
{
    return &func2;
}

statefn_t* func2()
{
    return &func1;
}

Answer 1

类型别名不能递归。

要实现诸如go讲座中使用的状态机，您需要定义自定义类型：

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());
}

Answer 2

正如Clearer评论的那样，这里是一个C ++类型stateFn的例子，它的行为类似于一个函数，并递归地返回相同类型的实例。

struct stateFn
{
    stateFn& operator() ();
}

Answer 3

如果你想在运行时解析递归，同时保持它与原始代码尽可能相似，你可以使用boost :: any或C ++ 17 std :: any，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 ) );
}

---

如果必须在编译时解析递归，则可以利用自动类型推导：

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()()();
}

当然，这在循环中不起作用，你需要某种编译时迭代方案才能使它有用......

Answer 4

这不是用C ++做事的自然方式。

在C ++中，它更像是混淆，这反映在为状态推进函数找到良好的自描述名称的困难中，这些函数也是状态：

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 );
    }
}