我对C ++的了解非常有限,甚至对fortran的了解也很少,我目前正在尝试从c ++ main调用fortran子例程。通过一些示例,我能够提出以下代码来调用
subroutine fireballess(ear,ne,parames,ifl,photar,photer)
这是我的C ++代码:
#include <stdlib.h>
#include <stdio.h>
#include <iostream>
using namespace std;
extern "C" void fireballess_( double *fear, int fne,double* fparames, int 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=[4.3,0.23,0.5,0.5,1.5,1.,1000.,2.15,3.]
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 = 2;
ear = new double[ne];
ear[0] = 0.;
ear[1] = 20.;
ear[2] = 40.;
ifl=2;
photar = new double[ne];
photer = new double[ne];
// Call a Fortran subroutine
//subroutine_sum_(&size,vec,&sum);
fireballess_(&ear,ne,¶mes,ifl,&photar,&photer);
cout << "Calling a Fortran subroutine" << endl;
cout << "===============================" << endl;
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;
}
但是,当我尝试编译时,出现以下错误:
call_fortran.cpp: In function ‘int main(int, char**)’:
call_fortran.cpp:45:53: error: cannot convert ‘double**’ to ‘double*’ for argument ‘1’ to ‘void fireballess_(double*, int, double*, int, double*, double)’
fireballess_(&ear,ne,¶mes,ifl,&photar,photer);
我不理解的,因为photer变量遵循photar变量的相同路径,所以不会产生任何错误。 我希望比我更了解指针的人可以在这里为我提供帮助。 谢谢
答案 0 :(得分:1)
ear
的类型为double*
,因此&ear
的类型为double**
,对于fortran函数原型来说,这是不合适的。摆脱&
的可能可以解决问题(&parames,&photar和&photer也是双重**):
更改
fireballess_(&ear,ne,¶mes,ifl,&photar,&photer);
到
fireballess_(ear,ne,parames,ifl,photar,photer);
fortran函数原型似乎是void fireballess_(double*, int, double*, int, double*, double)
。如果确实如此,则photer
仍必须更改为photer[i]
,其中(i是一个数组索引,如0、1、2 ...)。
如果您真的想了解发生了什么,我建议阅读一本不错的C ++指针教程。