Reputation: 583
At the moment I'm building functors (callable types) for the different calling conventions (__stdcall, __cdecl, __fastcall etc.). With the wrappers I'll be able to do something like this:
void __stdcall foo(int arg)
{
std::printf("arg: %i\n", arg);
}
int main(int, char**)
{
Function<void, int> v{foo};
v(1337);
return EXIT_SUCCESS;
}
At the moment I have built a wrapper for the __stdcall calling convention that is able to call any __stdcall function as long as the correct parameters are specified and the correct arguments are passed in. The class looks like this:
template <typename ReturnT, typename... Args>
class Function
{
// NOTE: This version of my callable types
// only supports the __stdcall calling
// convention. I need support for __cdecl,
// __fastcall and also __thiscall.
using return_t = ReturnT;
using callable_t = return_t(__stdcall*)(Args...);
private:
callable_t mCallable;
public:
template <typename FuncT>
Function(FuncT const &func) :
mCallable(func)
{
;
}
void operator()(Args&&... args)
{
mCallable(std::forward<Args>(args)...);
}
};
With that in hand I decided to build the other wrappers but I figured that typing out the same piece of code and changing the calling convention inside of the using declaration for callable_t is more work than needed. So I wanted to find a way to build about 4 variants of callable types (for each calling convention) but couldn't find a way to do it.
So far I have tried to use an enum as a non-type template parameter like this:
template <CallingConvention Call, typename ReturnT, typename... ArgsT>
class Function
{
// ...
};
But I don't know how to iterate the Call object's type and establish the required type (I tried utilizing std::is_same/std::enable_if but that was a dead end). I also tried template specialization with code like this:
struct StdcallT { ; };
struct CdeclT { ; };
struct FastcallT { ; };
template <typename CallT>
struct BaseT { };
template <> struct BaseT<StdcallT> { using CallableT = void(__stdcall*)(); };
template <> struct BaseT<CdeclT> { using CallableT = void(__cdecl*)(); };
template <> struct BaseT<FastcallT> { using CallableT = void(__fastcall*)(); };
template <typename CallT>
class Function
{
using CallableT = typename BaseT<CallT>::CallableT;
};
But I wasn't thinking of the rest of the arguments (return type + parameters) so this can't work too.
So does anyway have any ideas what I can do? One method I'm thinking of is doing a switch on the non-type parameter and calling the correct one like this:
template <CallingConvention Call, typename ReturnT, typename... ArgsT>
class Function
{
void operator()(ArgsT&&... args)
{
switch(Call)
{
case CallingConvention::Cdecl:
// Call a __cdecl version
break;
case CallingConvention::Stdcall:
// Call an __stdcall version
break;
// And so on...
}
}
};
And despite this looking like a working solution I was wondering if there was some good alternatives that I'm not thinking of.
Any ideas?
Upvotes: 8
Views: 1398
Reputation: 170193
Well once you define tags for each calling convention, you can use tag dispatch regularly:
#include <iostream>
#include <type_traits>
struct cdecl_tag { typedef void ( __attribute__((cdecl)) *type)(); };
struct stdcall_tag { typedef void ( __attribute__((stdcall)) *type)(); };
struct fastcall_tag { typedef void ( __attribute__((fastcall)) *type)(); };
constexpr void get_func_calling_convention_tag () {};
template<typename R, typename... Args>
constexpr cdecl_tag
get_func_calling_convention_tag (R (__attribute__((cdecl)) *)(Args...))
{ return {}; }
template<typename R, typename... Args>
constexpr stdcall_tag
get_func_calling_convention_tag (R (__attribute__((stdcall)) *)(Args...))
{ return {}; }
template<typename R, typename... Args>
constexpr fastcall_tag
get_func_calling_convention_tag (R (__attribute__((fastcall)) *)(Args...))
{ return {}; }
#define CALLING_CONVENTION_TAG(func) \
decltype(get_func_calling_convention_tag(&func))
int __attribute__((cdecl)) foo (char) { return 0; }
long __attribute__((stdcall)) bar (int) { return 0; }
int main()
{
std::cout << std::is_same<CALLING_CONVENTION_TAG(foo),
cdecl_tag>::value << '\n'
<< std::is_same<CALLING_CONVENTION_TAG(bar),
stdcall_tag>::value << '\n'
<< std::is_same<CALLING_CONVENTION_TAG(foo),
CALLING_CONVENTION_TAG(bar)>::value << std::endl;
return 0;
}
See it in action : http://ideone.com/HSZztX
This can of course be developed further; the tags may have a rebind variadic member template that returns a function pointer type with the appropriate calling convention specified.
I suppose you may even reduce the copying and pasting by having tag defintions neatly in a macro.
Upvotes: 1
Reputation: 679
If you still want to use the enumerated template argument, you can use specialization to accomplish this.
enum CallingConvention { __stdcall, ... };
template < CallingConvention Call >
struct implement {
template</* Template arguments for call method */>
static ReturnT call(/* arguments to run method */);
};
template < CallingConvention Call, typename ReturnT, typename... ArgsT >
class Function
{
// ...
template <typename FuncT>
Function(FuncT const &func) : mCallable(func), mCall(Call) {}
CallingConvention const mCall;
return_t operator()(ArgsT&&... args) {
return implement<Call>::call</* Template arguments for call method */>(/* arguments to run method */);
};
};
template < >
struct implement< __stdcall > {
template</* Template arguments for call method */>
static ReturnT call(/* arguments to run method */) {
// Special implementation...
}
};
That will be better than a switch statement.
(Sorry about the comments for the template arguments I am not quite familiar with how that works)
Here is where I got the idea for what I did.
Hope this helps!
Upvotes: 1