c++ - 如何编写用于在 C++14 中调用 Fortran 函数的通用包装器(按引用调用 --> 按值调用)

Question

我经常需要从我的 C++ 代码中调用一些 Fortran 例程。就我而言，C header 始终可用并包含诸如

之类的签名

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(可能使用模板元编程)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]));

应该等同于

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，但我不知道如何迭代一个参数包并在需要的地方引用。

Answer 1

对于这个答案，我将做出以下假设:

• FORTRAN 函数的参数都作为指针传递
• 指针地址将从传递给 fortranCall 函数的参数中获取。
• 数组指针参数后面总是跟着一个指向数组大小的指针
• 我们希望保留参数的顺序。

---

调用示例:

// 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);

---

以下程序将进行如上所示的调用:

#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。如果你想让它在 C++14 中工作，请使用来自 https://en.cppreference.com/w/cpp/utility/apply 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 ) 向后移植的调用