重新创建函数签名并通过C ++中的模板包进行调用

Question

I have C code that I want to rewrite in C++. The C code is part of a interpreter where the functions are defined in C however the actual call is from the interpreted source. Basically what it does is listed below:

#include <vector>

void f1(int a0) { }
void f2(int a0,int a1) { }
void f3(int a0,int a1,int a2) { }
void f4(int a0,int a1,int a2,int a3) { }

struct m {
    void *p;
    int c;
};

std::vector<m> ma;

int addfunc(void *p, int c) {
    int i = ma.size();
    ma.push_back({p,c});
    return i;
}

void call(int idx, int *stack) {
    switch (ma[idx].c) {
    case 1:
    ((void (*)(int))ma[idx].p)  (stack[0]);
    break;
    case 2:
    ((void (*)(int,int))ma[idx].p)  (stack[0],stack[1]);
    break;
    case 3:
    ((void (*)(int,int,int))ma[idx].p)  (stack[0],stack[1],stack[2]);
    break;
    case 4:
    ((void (*)(int,int,int,int))ma[idx].p)  (stack[0],stack[1],stack[2],stack[3]);
    break;
    }
}

int main (void) {
    int stack[5] = { 0,1,2,3,4 };
    /* define */
    int i1 = addfunc((void*)f1, 1);
    int i2 = addfunc((void*)f2, 2);
    int i3 = addfunc((void*)f3, 3);
    int i4 = addfunc((void*)f4, 4);
    /* call */
    call(i1,stack);
    call(i2,stack);
    call(i3,stack);
    call(i4,stack);
}

The addfunc creates a callable object specified by a function pointer and a signature, because the arguments are of the same type int only a count argument for the number of arguments is needed. When I call a function I specify the function object's index and a stack . The actual c-call is decoded via the argument count and typecasted, the call arguments are taken from the stack.

How can I rewrite the addfunc and call functions as templates objects in C++? How can I use template packs to count the number of arguments for the given function and regenerate the call to the function? How can I get rid of the switch statement and the function pointer typecast ? I have seen that luawrapper 's Binder class does something similar. However the code is quite complicated. In my case the arguments are all of the same type. In the end I want to do something like (pseudocode):

vector<meth> ma;
...
int i0 = addfunc([](int a) { });
int i1 = addfunc([](int a,int b) { });
int i2 = addfunc([](int a,int b,int b) { });
int i3 = addfunc([](int a,int b,int c,int c) { });
...
ma[i0](stack);
ma[i1](stack);
ma[i2](stack);
ma[i3](stack);

Answer 1

Well, if they're just C functions, why not overload on the function pointer type?

std::function<void(std::array<int, 5>)> addfunc(void (*f)(int)) {
    return [f](std::array<int, 5> const& a) { f(a[0]); };
}

std::function<void(std::array<int, 5>)> addfunc(void (*f)(int,int)) {
    return [f](std::array<int, 5> const& a) { f(a[0], a[1]); };
}

// repeat for all necessary arities

Then create std::vector<std::function<void(std::array<int, 5>)>> and push back all your functions. It's easy, doesn't require any templates and will work reasonably well. It introduces the overhead of std::function , though.

You could get rid of that by introducing your own callable type (n of them), that would correspond to the overloads above, provide an operator() and store appropriate function type inside.

Live example .

Answer 2

Unfortunately, you won't be able to make a completely generic solution, as there is no way to type-erase arity.

One way you can simplify things would be to create a set of wrappers for your functions, each wrapper accepting a stack* , and calling implementation functions with arguments from said stack .

Than you do not need typecasts at all and a simple function pointer (to approriate wrapper) would do (no even need to type-erase).

Answer 3

I propose a C++17 solution (simplified following a Jarod42's observation: thanks) that I suppose is over-complicated.

But I find it funny...

First: a struct that, given (as template parameters) a type and a unsigned number, define a type as the type received.

template <typename T, std::size_t>
struct getType
 { using type = T; };

It's used to convert a variadic template list of numbers in a sequence of types ( int s, in the following example) of the same length.

Next: a template type that register ( setFunc() ) and exec ( callFunc() ) a function returning void and a sequence of int s length as the first template parameter.

template <std::size_t N, typename = std::make_index_sequence<N>>
struct frHelper;

template <std::size_t N, std::size_t ... Is>
struct frHelper<N, std::index_sequence<Is...>>
 {
   using fnPnt_t = void(*)(typename getType<int, Is>::type...);

   fnPnt_t fp = nullptr;

   void setFunc (fnPnt_t fp0)
    { fp = fp0; }

   void callFunc (std::array<int, sizeof...(Is)> const & a)
    { if ( fp ) fp(a[Is]...); }
 };

Last: a template struct that inherit from a variadic list of preceding structs and enable ( using ) the setFunc() and the callFunc() members.

template <std::size_t N, typename = std::make_index_sequence<N>>
struct funcRegister;

template <std::size_t N, std::size_t ... Is>
struct funcRegister<N, std::index_sequence<Is...>>
   : public frHelper<Is>...
 { 
   using frHelper<Is>::setFunc...;
   using frHelper<Is>::callFunc...;
 };

Use.

First you have to declare an object of type funcRegister<N> where N is the max number of integer received from your functions plus one. So if you want to use f4() , so four integers, you have to declare

   funcRegister<5u>  fr;

Then you have to register the functions

   fr.setFunc(f1);
   fr.setFunc(f2);
   fr.setFunc(f3);
   fr.setFunc(f4);

and, given some std::array<int, N> of the right size, you can call the registered functions

   std::array a1 { 1 };
   std::array a2 { 1, 2 };
   std::array a3 { 1, 2, 3 };
   std::array a4 { 1, 2, 3, 4 };

   fr.callFunc(a1); // call f1
   fr.callFunc(a2); // call f2
   fr.callFunc(a3); // call f3
   fr.callFunc(a4); // call f4

The following is a full compiling C++17 example

#include <array>
#include <utility>
#include <iostream>
#include <type_traits>

template <typename T, std::size_t>
struct getType
 { using type = T; };

template <std::size_t N, typename = std::make_index_sequence<N>>
struct frHelper;

template <std::size_t N, std::size_t ... Is>
struct frHelper<N, std::index_sequence<Is...>>
 {
   using fnPnt_t = void(*)(typename getType<int, Is>::type...);

   fnPnt_t fp = nullptr;

   void setFunc (fnPnt_t fp0)
    { fp = fp0; }

   void callFunc (std::array<int, sizeof...(Is)> const & a)
    { if ( fp ) fp(a[Is]...); }
 };

template <std::size_t N, typename = std::make_index_sequence<N>>
struct funcRegister;

template <std::size_t N, std::size_t ... Is>
struct funcRegister<N, std::index_sequence<Is...>>
   : public frHelper<Is>...
 { 
   using frHelper<Is>::setFunc...;
   using frHelper<Is>::callFunc...;
 };

void f1(int) { std::cout << "f1 called" << std::endl; }
void f2(int,int) { std::cout << "f2 called" << std::endl;}
void f3(int,int,int) { std::cout << "f3 called" << std::endl;}
void f4(int,int,int,int) { std::cout << "f4 called" << std::endl;}

int main()
 {
   funcRegister<5u> fr;

   fr.setFunc(f1);
   fr.setFunc(f2);
   fr.setFunc(f3);
   fr.setFunc(f4);

   std::array a1 { 1 };
   std::array a2 { 1, 2 };
   std::array a3 { 1, 2, 3 };
   std::array a4 { 1, 2, 3, 4 };

   fr.callFunc(a1);
   fr.callFunc(a2);
   fr.callFunc(a3);
   fr.callFunc(a4);    
 }

Answer 4

Here is luawrapper 's code extracted to be applied the above case. This is more for completion as for @Jerod42's code is preferable.

#include <iostream>
#include <string>
#include <array>
#include <vector>
#include <functional>
#include <vector>

template<typename T> struct tag {};

template<typename TFunctionObject, typename TFirstParamType>
struct Binder {
    TFunctionObject function;
    TFirstParamType param;

    template<typename... TParams>
    auto operator()(TParams&&... params)
    -> decltype(function(param, std::forward<TParams>(params)...))
    {
        return function(param, std::forward<TParams>(params)...);
    }
};

template<typename TCallback>
static void readIntoFunction(int *stack, TCallback&& callback)
{
    callback();
}
template<typename TCallback, typename TFirstType, typename... TTypes>
static void readIntoFunction(int *stack, TCallback&& callback, tag<TFirstType>, tag<TTypes>... othersTags)
{
    Binder<TCallback, const TFirstType&> binder{ callback, *stack };
    return readIntoFunction(++stack, binder, othersTags...);
}

/* decompose arguments */
template<typename TFunctionType, typename... TOtherParams>
std::function<void(int*)> _addfunc(TFunctionType f, tag<void (*)(TOtherParams...)>) {
    return std::function<void(int*)>([f](int *stack) {
        readIntoFunction(stack, f, tag<TOtherParams>{}...);
    });
}

template<typename TFunctionType>
std::function<void(int*)> addfunc(TFunctionType fn)
{
    typedef typename std::decay<TFunctionType>::type RealFuncSig;
    return _addfunc(std::move(fn), tag<RealFuncSig>{} );
}

void f1(int a0) { std::cout << a0 << std::endl; }
void f2(int a0, int a1) { std::cout << a0 << a1 << std::endl;  }

int main() {
    int stack[5] = { 0,1,2,3,4 };
    auto a0 = addfunc(&f1);
    auto a1 = addfunc(&f2);

    a0(stack);
    a1(stack);
}

Answer 5

您可以使用std：function作为addfun（）的参数，也可以使用std :: bind