Question

我试图在C ++中调用Fortran子例程。

这是子例程的开始：

    subroutine fireballess(ear,ne,parames,ifl,photar,photer)
    integer ne,ifl
    real*4 ear(0:ne),parames(10),photar(ne),photer(ne)

该子例程在Fortran中工作正常，但是当我尝试在C ++中调用它时，出现了段错误。这是我的代码：

    #include <stdlib.h>
    #include <stdio.h>
    #include <iostream>

    using namespace std;

    extern "C" void fireballess_( double *fear, int fne,double* fparames, nt fifl, double *fphotar, double *fphoter);

    int main(int argc, char ** argv)
    {
            int ne,ifl;
            double *ear;
        double *parames;

        double *photar;
            double *photer;


        parames = new double[9];

            parames[0]=4.3;
           parames[1]=0.23;
           parames[2]=0.5;
           parames[3]=0.5;
           parames[4]=1.5;
           parames[5]=1.;
           parames[6]=1000.;
           parames[7]=2.15;
           parames[8]=3.;

        ne = 3;

            ear = new double[ne];
        ear[0] = 0.;
        ear[1] = 20.;
        ear[2] = 40.;
        ifl=2;

            photar = new double[ne];
            photer = new double[ne];

    // Check on variables initialization

        for (int i=0;i<=2;i++)    cout << ear[i] <<",";
        cout <<"    "<< ne<<"   ";
        for (int i=0;i<=8;i++)    cout << parames[i] <<",";
        cout <<"    "<< ifl <<" "<< photar[0] <<"   "<< photer[0] << endl;

            cout << "Calling a Fortran subroutine" << endl;
            cout << "===============================" << endl;

    // call to the subroutine -->segmentation fault
            fireballess_(ear,ne,parames,ifl,photar,photer);

        for (int i=0;i<=ne;i++){
        cout << "ear = " <<ear[i-1]<< " - "<<ear[i] << endl;
        cout << "photar = " << photar[i] << endl;
        cout << "photer = " << photer[i] << endl << endl;
        }

            delete[] ear;
            delete[] parames;
            delete[] photar;
            delete[] photer;
    }

程序在调用子例程时崩溃。我不太熟悉C ++或Fortran编码，所以我不确定该怎么做。到目前为止，我已经检查了传递给子例程的变量的格式是否正确，看起来是正确的。

在此先感谢您的帮助

---------编辑----- 阅读一些注释后，我对代码进行了如下修改，但在调用例程时仍然遇到相同的分段错误错误：

using namespace std;

extern "C" void fireballess_( std::vector<float> fear, int fne,std::vector<float> fparames, int fifl, std::vector<float> fphotar, std::vector<float> fphoter);

int main(int argc, char ** argv)
{
    int ne,ifl;
ifl=2;
ne = 3;
    std::vector<float> parames = {4.3,0.23,0.5,0.5,1.5,1.,1000.,2.15,3.};
std::vector<float> ear={0,20,40};
    std::vector<float> photar;
    std::vector<float> photer;

    cout << "Calling a Fortran subroutine" << endl;
    cout << "===============================" << endl;
        fireballess_(ear,ne,parames,ifl,photar,photer);

for (int i=0;i<ne;i++){
        cout << "ear = " <<ear[i-1]<< " - "<<ear[i] << endl;
        cout << "photar = " << photar[i] << endl;
        cout << "photer = " << photer[i] << endl << endl;
}

}

Answer 1

结合注释和其他答案中的各种信息，我想原始代码的最低修改版本可能看起来像这样：

fortsub.f90

subroutine fireballess(ear, ne, parames, ifl, photar, photer)
    implicit none
    integer ne, ifl
    real*4 ear(0:ne), parames(10), photar(ne), photer(ne)

    print *, "ear     = ", ear
    print *, "ne      = ", ne
    print *, "parames = ", parames
    print *, "ifl     = ", ifl
    print *, "photar  = ", photar
    print *, "photer  = ", photer
    print *
    print *, "sizeof(integer) = ", sizeof(ne)
    print *, "sizeof(real*4)  = ", sizeof(ear(1))
end

main.cpp

#include <iostream>

extern "C"
void fireballess_( float *ear, int *ne, float *parames,
                   int *ifl, float *photar, float *photer );

int main()
{
    std::cout << "sizeof(int)   = " << sizeof(int)   << std::endl;
    std::cout << "sizeof(float) = " << sizeof(float) << std::endl;

    int ne = 3, ifl = 2;

    float *ear = new float[ne + 1] { 0.0f, 20.0f, 40.0f, 60.0f };

    float *parames = new float[10]
        { 0.1f, 0.2f, 0.3f, 0.4f, 0.5f,
          0.6f, 0.7f, 0.8f, 0.9f, 1.0f };

    float *photar = new float[ne] { 1.0f, 2.0f, 3.0f };
    float *photer = new float[ne] { 4.0f, 5.0f, 6.0f };

    fireballess_( ear, &ne, parames, &ifl, photar, photer );

    delete[] ear;
    delete[] parames;
    delete[] photar;
    delete[] photer;
}

在这里，修改后的部分是：（1）Fortran中的real*4通常对应于C ++中的float；（2）我们需要传递实际参数的地址（尤其是ne和ifl的地址）；（3）ear应该具有ne + 1元素，因为在Fortran端它被定义为ear(0:ne)。然后，将代码编译为

$ gfortran-8 -c fortsub.f90
$ g++-8 main.cpp fortsub.o -lgfortran

提供（使用GCC8 + MacOS10.13）

sizeof(int)   = 4
sizeof(float) = 4
 ear     =    0.00000000       20.0000000       40.0000000       60.0000000    
 ne      =            3
 parames =   0.100000001      0.200000003      0.300000012      0.400000006      0.500000000      0.600000024      0.699999988      0.800000012      0.899999976       1.00000000    
 ifl     =            2
 photar  =    1.00000000       2.00000000       3.00000000    
 photer  =    4.00000000       5.00000000       6.00000000    

 sizeof(integer) =                     4
 sizeof(real*4)  =                     4

如果我们使用std::vector，则相应的代码可能如下所示（通过查看有关std::vector的一些教程...）

// main2.cpp
#include <iostream>
#include <vector>

extern "C"
void fireballess_( float *ear, int *ne, float *parames,
                   int *ifl, float *photar, float *photer );

using Vecf = std::vector<float>;

int main()
{
    int ne = 3, ifl = 2;

    Vecf ear { 0.0f, 20.0f, 40.0f, 60.0f };

    Vecf parames { 0.1f, 0.2f, 0.3f, 0.4f, 0.5f,
                   0.6f, 0.7f, 0.8f, 0.9f, 1.0f };

    Vecf photar { 1.0f, 2.0f, 3.0f };
    Vecf photer { 4.0f, 5.0f, 6.0f };

    fireballess_( ear.data(), &ne, parames.data(),
                  &ifl, photar.data(), photer.data() );
}

似乎给出了相同的结果（通过将main.cpp替换为main2.cpp）。

Answer 2

从C和C ++代码调用Fortran函数的现代方法是通过ISO_C_BINDING，请参见tag wiki on SO。如果没有此界面，就会出现一些类似的问题

所有变量都通过引用（指针）传递给Fortran函数
传递数组时，还必须传递其长度
...

C / Fortran互操作这种古老的古怪方式在例如here。

请注意，您不能将std::vector或任何其他复杂的C ++对象传递给C或Fortran函数，并期望它会理解。除非该函数旨在处理此问题，否则这很可能行不通。因此，您需要从这些对象中获取实际的低级数据（例如，指向数组的指针），并改为传递它们。

C ++中调用fortran子例程时出现分段错误

2 个答案: