Question

通常，我必须从我的C ++代码中调用一些Fortran例程。就我而言，C标头始终可用，并且包含诸如

double fFortran(int* a, int* b, double* someArray, int* sizeOfThatArray)

我的问题是：是否可以编写通用的C ++ 14包装器fortranCall（也许使用模板元编程）它会在必要时获取地址，然后调用fortran函数像这样

double someArray[2] = {1, 4};
double result = fortranCall(fFortran, 4, 5, someArray,
    sizeof(someArray) / sizeof(someArray[0]));

应等于

double someArray[2] = {1, 4};
int sizeOfSomeArray = sizeof(someArray) / sizeof(someArray[0]);
int a = 4;
int b = 5;
double result = fFortran(&a, &b, someArray, &sizeOfSomeArray);

我认为正确的解决方案涉及参数包，但我无法弄清楚如何迭代一个参数包并在需要时获取引用。

Answer 1

对于这个答案，我将做以下假设：

FORTRAN函数的参数都作为指针传递
要从传递给fortranCall函数的参数中获取指针地址。
数组指针参数后面总是一个指向数组大小的指针
我们要保留参数的顺序。

示例调用：

// So, given function signature
double fFortran(int* a, int* b, double* someArray, int* sizeOfThatArray);
// we would like to call with:
fortranCall(fFortran, 4, 5, someArray);

// Likewise, given
fFortranTwoArrays(double* arrayA, int* size_of_A, double* arrayB, int* size_of_B);
// we would like to call with
fortranCall(fFortranTwoArrays, someArray, some_other_Array);

以下程序将进行调用，如上所示：

#include <tuple>
#include <type_traits>

// Functions to call eventually
double fFortran(int* a, int* b, double* someArray, int* sizeOfThatArray)
{ 
    return 0.0; 
}

double fFortranTwoArrays(double* arrayA, int* size_of_A, double* arrayB, int* size_of_B)
{ 
    return 0.0; 
}

// If T is an array 
// then make a std::tuple with two parameters
//   pointer to first of T and 
//   pointer to extent of T
template<
    typename T,
    typename std::enable_if <
        std::is_array<T>{},
        int
    >::type Extent = std::extent<T>::value,
    typename Ptr = typename std::decay<T>::type
>
auto make_my_tuple(T& t)
{
    static auto extent = Extent;
    Ptr ptr = &t[0];
    return std::make_tuple(ptr, &extent);
}

// If T is not an array 
// then make a std::tuple with a single parameter
//   pointer to T
template<typename T,
    typename std::enable_if <
        !std::is_array<T>{},
        int
    >::type = 0 
>
auto make_my_tuple(T& t)
{
    return std::make_tuple(&t);
}

template<typename F, typename... Targs>
auto fortranCall(F& f, Targs&& ... args)
{
    // Make a single tuple with all the parameters.
    auto parameters = std::tuple_cat(make_my_tuple(args)...);

    // Arrays were each expanded to 
    // two pointer parameters(location and size).
    // Other parameters will pass as a single pointer
    return std::apply(f,parameters);
}

int main()
{
    double someArray[2] = {1, 4};
    double result = fortranCall(fFortran, 4, 5, someArray);

    double some_other_Array[] = {6,7,8,9,10};
    auto result2 = fortranCall(fFortranTwoArrays, someArray, some_other_Array);
}

std :: apply是C ++ 17。如果要使其在C ++ 14中工作，请使用https://en.cppreference.com/w/cpp/utility/apply

中的示例实现

namespace detail {
template <class F, class Tuple, std::size_t... I>
constexpr decltype(auto) apply_impl(F&& f, Tuple&& t, std::index_sequence<I...>)
{
    return std::invoke(std::forward<F>(f), std::get<I>(std::forward<Tuple>(t))...);
}
}  // namespace detail

template <class F, class Tuple>
constexpr decltype(auto) apply(F&& f, Tuple&& t)
{
    return detail::apply_impl(
        std::forward<F>(f), std::forward<Tuple>(t),
        std::make_index_sequence<std::tuple_size<std::remove_reference_t<Tuple>>::value>{});
}

并使用Martin Moene（https://github.com/martinmoene/invoke-lite）从后台的调用

如何编写通用包装程序以在C ++ 14中调用Fortran函数（按引用调用->按值调用）

1 个答案: