C++11可变模板函数调用转发

C++11 variadic template function call forwarding

本文关键字：函数调用 转发 C++11      更新时间：2023-10-16

我正在尝试创建一个C++11模板函数，该函数将在两种约定之间转换函数调用：第一种是使用Variant(注意：Variant是多态类型，是IntVariable、DoubleVariant等子类的基础)，第二种是C函数调用。

我们在编译时知道每一条信息：参数计数是参数的数量，参数/返回类型取决于"cfunc"变量类型。

// We will assume that the two following functions are defined with their correct
// specializations.
template < typename T >
Variant * convertToVariant( T t );
template < typename T >
T convertFromVariant( Variant * variant );
// The following function is incomplete, the question is how to convert the
// variant parameters into a C function call ?
template < typename Return, typename... Arguments >
Variant * wrapCFunction< Return cfunc( Args... ) >(int argc, Variant ** argv) {
// Here comes the magic call of cfunc, something like :
if ( argc != mpl::count< Args... >::value )
throw std::runtime_error( "bad argument count" );
return cfunc( convertFromVariant< Args... >( argv[ X ] )... );
}
// Example use case :
int foo( int a, int b );
int main(void) {
int argc = 2;
Variant * argv[2] = { new IntVariant( 5 ), new IntVariant( 6 ) };
Variant * res = wrapCFunction< foo >( argc, argv );
IntVariant * intRes = dynamic_cast< IntVariant >( res );
return intRes ? intRes->value : -1;
}

使用索引技巧，这很容易：

template<unsigned...> struct indices{};
template<unsigned N, unsigned... Is>
struct indices_gen : indices_gen<N-1, N-1, Is...>{};
template<unsigned... Is>
struct indices_gen<0, Is...> : indices<Is...>{};
// assuming the parameters were actually like this
template<typename Return, typename... Args, unsigned... Is>
Variant* wrapCFunction(Return (*cfunc)(Args...), int argc, Variant** argv, indices<Is...>) {
return cfunc(convertFromVariant<Args>(argv[Is])...);
}
template<typename Return, typename... Args>
Variant* wrapCFunction(Return (*cfunc)(Args...), int argc, Variant** argv) {
if (argc != sizeof...(Args))
throw std::runtime_error("bad argument count");
return wrapCFunction(cfunc, argc, argv, indices_gen<sizeof...(Args)>());
}

请注意代码中的一些更改。首先，sizeof...(Args)生成包中的参数数量。其次，我修复了函数的签名，以传递cfunc作为实际参数。

class Variant;
template < typename T > Variant * convertToVariant( T t );
template < typename T > T convertFromVariant( Variant * variant );
template <typename Return, typename... Arguments>
struct WrapCFunctionImpl {
template<int argsToAdd, int... argIndexes>
struct Impl {
typedef typename Impl<argsToAdd - 1, argIndexes..., sizeof...(argIndexes)>::Type Type;
};
};
template <typename Return, typename... Arguments>
template <int... argIndexes>
struct WrapCFunctionImpl<Return, Arguments...>::Impl<0, argIndexes...> {
typedef Impl Type;
static Variant* run(Return cfunc( Arguments... ), Variant ** argv) {
return convertToVariant( cfunc( convertFromVariant<Arguments>( argv[argIndexes] )... ) );
}
};
template < typename Return, typename... Arguments >
Variant * wrapCFunction(Return cfunc( Arguments... ), Variant ** argv) {
return WrapCFunctionImpl<Return,Arguments...>::template Impl<sizeof...(Arguments)>::Type::run(cfunc, argv);
}
int foo(int, int);
Variant *f(Variant** x) {
return wrapCFunction(foo, x);
}

这里的大部分困难是递归生成数组中的索引。也许有更简单的方法，但这是有效的。

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