【发布时间】：2016-03-09 01:03:53
【问题描述】：
我希望从我的 C++ 代码中调用 fortran 例程 cbesj.f，我该如何实现？
以下是我已完成的步骤：
从 netlib amos 网页下载 cbesj.f 以及依赖项，http://www.netlib.org/cgi-bin/netlibfiles.pl?filename=/amos/cbesj.f
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在源目录中，
f2c -C++PR *.f
g++ -c *.c
ar cr libmydemo.a *.o
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[test_cbesj.cpp][1] 和 [mydemo.h][2]就是这样调用子程序的，
g++ test_cbesj.cpp -lf2c -lm -L。 -lmydemo 它返回错误：
test_cbesj.cpp:(.text+0xd6): 未定义引用 `cbesj_(complex*, float*, long*, long*, complex*, long*, long*)'
在我的问题中引用 cbesj_ 子程序的正确方法是什么？谢谢！
感谢凯西： 我认为你的方法是最好的。但是我还是设置错了，为什么？我们开始：
f77 -c *.f
在 modemo.h 中
//File mydemo.h
#ifndef MYDEMO_H
#define MYDEMO_H
#include <stdio.h> /* Standard Library of Input and Output */
#include "f2c.h"
extern"C" int cacai_(complex *z__, real *fnu, integer *kode, integer *mr, integer *n, complex *y, integer *nz, real *rl, real *tol, real *el\
im, real *alim);
extern"C" int cairy_(complex *z__, integer *id, integer *kode, complex *ai, integer *nz, integer *ierr);
extern"C" int casyi_(complex *z__, real *fnu, integer *kode, integer *n, complex *y, integer *nz, real *rl, real *tol, real *elim, real \
*alim);
extern"C" int cbesj_(complex *z__, real *fnu, integer *kode, integer *n, complex *cy, integer *nz, integer *ierr);
extern"C" int cbinu_(complex *z__, real *fnu, integer *kode, integer *n, complex *cy, integer *nz, real *rl, real *fnul, real *tol, real *el\
im, real *alim);
extern"C" int cbknu_(complex *z__, real *fnu, integer *kode, integer *n, complex *y, integer *nz, real *tol, real *elim, real *alim);
extern"C" int cbuni_(complex *z__, real *fnu, integer *kode, integer *n, complex *y, integer *nz, integer *nui, integer *nlast, real *fnul, \
real *tol, real *elim, real *alim);
extern"C" int ckscl_(complex *zr, real *fnu, integer *n, complex *y, integer *nz, complex *rz, real *ascle, real *tol, real *elim);
extern"C" int cmlri_(complex *z__, real *fnu, integer *kode, integer *n, complex *y, integer *nz, real *tol);
extern"C" int crati_(complex *z__, real *fnu, integer *n, complex *cy, real *tol);
extern"C" int cs1s2_(complex *zr, complex *s1, complex *s2, integer *nz, real *ascle, real *alim, integer *iuf);
extern"C" int cseri_(complex *z__, real *fnu, integer *kode, integer *n, complex *y, integer *nz, real *tol, real *elim, real *alim);
extern"C" int cshch_(complex *z__, complex *csh, complex *cch);
extern"C" int cuchk_(complex *y, integer *nz, real *ascle, real *tol);
extern"C" int cunhj_(complex *z__, real *fnu, integer *ipmtr, real *tol, complex *phi, complex *arg, complex *zeta1, complex *zeta2, complex\
*asum, complex *bsum);
extern"C" int cuni1_(complex *z__, real *fnu, integer *kode, integer *n, complex *y, integer *nz, integer *nlast, real *fnul, real *tol \
, real *elim, real *alim);
extern"C" int cuni2_(complex *z__, real *fnu, integer *kode, integer *n, complex *y, integer *nz, integer *nlast, real *fnul, real *tol \
, real *elim, real *alim);
extern"C" int cunik_(complex *zr, real *fnu, integer *ikflg, integer *ipmtr, real *tol, integer *init, complex *phi, complex *zeta1, complex\
*zeta2, complex *sum, complex *cwrk);
extern"C" int cuoik_(complex *z__, real *fnu, integer *kode, integer *ikflg, integer *n, complex *y, integer *nuf, real *tol, real *elim, re\
al *alim);
extern"C" int cwrsk_(complex *zr, real *fnu, integer *kode, integer *n, complex *y, integer *nz, complex *cw, real *tol, real *elim, real *a\
lim);
extern"C" real gamln_(real *z__, integer *ierr);
extern"C" integer i1mach_(integer *i__);
extern"C" real r1mach_(integer *i__);
extern"C" int xerror_(char *mess, integer *nmess, integer *l1, integer *l2, ftnlen mess_len);
#endif
在 test_cbesj.cpp 中，
#include "mydemo.h"
#include "f2c.h"
#include <math.h>
#include <iostream>
#include <stdio.h>
#include <assert.h>
using namespace std;
int main(void)
{
// double x=86840.;
//int nu=46431,j, err;
complex *z,z__;
z__.r = 3.0;z__.i = 2.0;z = &z__;
cout << z->r << '\t' << z->i << endl;
real *fnu;float fnu__ = 3.0;fnu = &fnu__;
integer *kode ;long int kode__=1;kode=&kode__;
integer *n ;long int n__=1;n=&n__;
complex *cy;
integer *nz;
integer *ierr;
cbesj_(z, fnu, kode, n, cy, nz, ierr);
cout << cy->r << '\t' << cy->i << endl;
return 0;
}
那么，
g++ -c -g test_cbesj.cpp
g++ -o test *.o -lg2c
./test
3 2
Segmentation fault (core dumped)
gdb test
GNU gdb (Ubuntu/Linaro 7.4-2012.04-0ubuntu2.1) 7.4-2012.04
Copyright (C) 2012 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law. Type "show copying"
and "show warranty" for details.
This GDB was configured as "i686-linux-gnu".
For bug reporting instructions, please see:
<http://bugs.launchpad.net/gdb-linaro/>...
Reading symbols from /media/Downloads/amos-4/test...done.
(gdb) run
Starting program: /media/Downloads/amos-4/test
3 2
Program received signal SIGSEGV, Segmentation fault.
0x0804b355 in cbesj_ ()
(gdb) frame 0
#0 0x0804b355 in cbesj_ ()
(gdb) frame 1
#1 0x0805a3ca in main () at test_cbesj.cpp:21
21 cbesj_(z, fnu, kode, n, cy, nz, ierr);
感谢 roygvib 的回复！其实很好的建议。这是更改后的 test_cbesj.cpp：
complex z, cy;
float fnu;
long int kode, n, nz, ierr;
z.r = 3.0; z.i = 2.0;
fnu = 3.0;
n = 1; kode = 1;
cout.precision(16);
cbesj_( &z, &fnu, &kode, &n, &cy, &nz, &ierr );
cout << cy.r << '\t' << cy.i << endl;
cout << "nz=" << nz << endl;
cout << "ierr=" << ierr << Lendl;
没有更多的段错误。但是由于某些原因，代码没有按预期工作：
./test
-1.343533039093018 -1.343533992767334
nz=0
ierr=4
而且答案是错误的，ierr 从源代码中也这么说：
C NZ - NUMBER OF COMPONENTS SET TO ZERO DUE TO UNDERFLOW,
C NZ= 0 , NORMAL RETURN
C NZ.GT.0 , LAST NZ COMPONENTS OF CY SET TO ZERO
C DUE TO UNDERFLOW, CY(I)=CMPLX(0.0,0.0),
C I = N-NZ+1,...,N
C IERR - ERROR FLAG
C IERR=0, NORMAL RETURN - COMPUTATION COMPLETED
C IERR=1, INPUT ERROR - NO COMPUTATION
C IERR=2, OVERFLOW - NO COMPUTATION, AIMAG(Z)
C TOO LARGE ON KODE=1
C IERR=3, CABS(Z) OR FNU+N-1 LARGE - COMPUTATION DONE
C BUT LOSSES OF SIGNIFCANCE BY ARGUMENT
C REDUCTION PRODUCE LESS THAN HALF OF MACHINE
C ACCURACY
C IERR=4, CABS(Z) OR FNU+N-1 TOO LARGE - NO COMPUTA-
C TION BECAUSE OF COMPLETE LOSSES OF SIGNIFI-
C CANCE BY ARGUMENT REDUCTION
C IERR=5, ERROR - NO COMPUTATION,
C ALGORITHM TERMINATION CONDITION NOT MET
【问题讨论】：
-
你有没有检查过
f2c生成的任何C文件中是否定义了函数？ -
另外，请发布您链接到的文件的内容。这里强烈建议不要链接到源代码。当链接指向图像时，情况会更糟。
-
至少，你必须让你的头文件将函数声明为
extern C，这样它正在寻找的函数名就不会被破坏。 -
搜索关于 fortran 与 c 的互操作性主题的 Qs 和 As，了解如何在不通过 f2c 的情况下从 C++ 调用 Fortran 例程。
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使 Fortran 可互操作。几乎可以肯定，使用 f2c 编译会对代码的性能产生负面影响。使用真正的 Fortran 编译器并将所有内容链接在一起。搜索iso_c_binding，这里有很多例子。
标签： c++ fortran fortran77 f2c netlib