在 C++ 中调用 fortran 子程序时出现分段错误 [英] segmentation fault when calling fortran subroutine in C++
问题描述
我正在尝试在 C++ 中调用 Fortran 子例程.
I am trying to call a Fortran subroutine in C++.
这是子程序开始:
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 中运行良好，但是当我尝试在 C++ 中调用它时，出现分段错误.这是我的代码:
The subroutine works fine in Fortran, but when I try to call it in C++ I get a segmentation fault. Here is my code:
#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.我在 C++ 或 Fortran 编码方面不是很有经验，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程序的变量的格式是否正确，看起来是这样.
The program crashes on the call to the subroutine. I am not very experienced in C++ or Fortran coding, so I am not sure about what to do. So far I have checked the format of the variables passed to the subroutine was correct, and it looks so.
在此先感谢您的帮助
---------编辑------在阅读了一些评论后，我将代码修改如下，调用例程时仍然出现相同的分段错误错误:
---------EDIT------ after reading some comments I revised the code as follows, still getting the same segmentation fault error when the routine is invoked:
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;
}
}
推荐答案
结合评论和其他答案中的各种信息，我猜原始代码的最小修改版本可能如下所示:
Combining various info in the comments and the other answer, I guess a minimum-modified version of the original code may look like this:
fortsub.f90
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;
}
这里修改的部分是: (1) Fortran中的real*4通常对应C++中的float；(2)我们需要传递实参的地址(特别是ne和ifl)；(3) ear 应该有 ne + 1 元素，因为它在 Fortran 端被定义为 ear(0:ne) .然后，将代码编译为
Here, the modified parts are: (1) real*4 in Fortran usually corresponds to float in C++; (2) we need to pass the address of actual arguments (particularly for ne and ifl); and (3) ear should have ne + 1 elements because it is defined as ear(0:ne) on the Fortran side. Then, compiling the codes as
$ gfortran-8 -c fortsub.f90
$ g++-8 main.cpp fortsub.o -lgfortran
给出(使用 GCC8 + MacOS10.13)
gives (with 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，对应的代码可能是这样的(通过查看一些关于std::vector...的教程)
If we use std::vector, the corresponding code may look like this (by looking at some tutorials on std::vector...)
// 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() );
}
这似乎给出了相同的结果(通过将 main.cpp 替换为 main2.cpp).
which seems to give the same result (by replacing main.cpp by main2.cpp).
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