Fortran接受来自C的字符串（？） [英] Fortran accepting string(?) from C

问题描述

我觉得这应该是一个简单的问题，但我无法实现它的工作。我有一些Fortran代码需要输入：

  SUBROUTINE TRACE（X，Y，NAME，XX，YY）
 EXTERNAL NAME 
 CALL NAME（X，Y，XX，YY）
  

和I 'm试图从C ++传入一个名字，形式如下：

  float x，y，xx，yy; 
 char * name =IGRF; 
 trace _（&x，&y，name，&xx，&yy）; 
  

它会编译，但当我尝试调用NAME子例程时，我总是会遇到段错误。在文件中定义了一个名为IGRF的子例程，我可以直接从C ++调用IGRF子例程，但需要此TRACE例程。在gdb中运行时，它表示NAME变量作为指向void的指针传递。

我试过传递NAME，& NAME和& NAME [0] ，一个字符NAME [4]被剥离了\0以完全符合该名称，并且他们都回来显示相同​​的空指针。是否有人知道如何从C ++获取函数名称到Fortran中的EXTERNAL变量？

谢谢

解决方案

因此，Fortran2003及其后续版本的一个优点是将C互操作性定义纳入标准;这是一个PITA的使用，但一旦完成，它将保证跨平台和编译器工作。

所以这里是 cprogram.c ，调用Fortran例程 getstring ：

  #include< stdio.h> 
 
 int main（int argc，char ** argv）{
 int l; 
 char * name =IGRF; 
 
 l = getstring（name）; 
 
 printf（In C：l =％d \\\
，l）; 
 
返回0; 
} 
  

以下是 fortranroutine.f90 ：

 整数（kind = c_int）函数getstring（instr）bind（C，name ='getstring'）
 use，intrinsic :: iso_c_binding 
 character（kind = c_char），dimension（*），intent（IN）:: instr 
 integer :: len 
 integer :: i $ b $如果（instr（len + 1）== C_NULL_CHAR）退出
 len = len + 1 
 end do 
 
 $ b len = 0 
 do 
 b'b 
 print *，'在Fortran中：'
 print *，'Got string：'，（instr（i），i = 1，len）
 getstring = len 
结束函数getstring 
  

makefile非常简单：

  CC = gcc 
 FC = gfortran 
 
 cprogram：cprogram.o fortranroutine.o 
 $（CC）-o cprogram cprogram.o fortranroutine.o -lgfortran 
 
 fortranroutine.o：fortranroutine.f90 
 $（FC）-c $ ^ 
 
 clean：
 rm -f * .o cprogram *〜
  

并在gcc / gfortran和icc / ifort下运行它：
$ b

 在Fortran中：
字符串：IGRF 
在C：l = 4 
  

更新：哦，我刚刚意识到你正在做的事情比传递一个字符串更精细;你基本上试图传递一个指向C回调例程的函数指针。这是一个小窍门，因为你必须使用Fortran interface 来声明C例程 - 只是使用extern不起作用（并且不如显式接口无论如何，因为没有类型检查等）所以这应该工作：

cprogram.c：

  #include< stdio.h> 
 
 / * fortran例程原型* / 
 int getstring（char * name，int（*）（int））; 
 
 int square（int i）{
 printf（In C called from Fortran :,）; 
 printf（％d平方是％d！\\\
，i，i * i）; 
返回i * i; 
 
 
 
 int cube（int i）{
 printf（In C called from Fortran :,）; 
 printf（％d cubed is％d！\\\
，i，i * i * i）; 
返回i * i * i; 
 
 
 int main（int argc，char ** argv）{
 int l; 
 char * name =IGRF; 
 
 l = getstring（name，& square）; 
 printf（In C：l =％d \\\
，l）; 
 l = getstring（name，& cube）; 
 printf（In C：l =％d \\\
，l）; 
 
 
 return 0; 
 
  

froutine.f90：

  integer（kind = c_int）函数getstring（str，func）bind（C，name ='getstring'）
 use，intrinsic :: iso_c_binding 
 implicit none 
 character（kind = c_char），dimension（*），intent（in）:: str 
 type（c_funptr），value :: func 
 
 integer :: length 
 integer :: i 
 
！ C函数的原型;取一个c_int，返回一个c_int 
接口
整数（kind = c_int）function croutine（inint）bind（C）
 use，intrinsic :: iso_c_binding 
 implicit none 
整数（kind = c_int），value :: inint 
结束函数croutine 
结束接口
过程（croutine），指针:: cfun 
 
整数（kind = c_int）:: clen 
 
！将C转换为fortran程序指针
！与称为croutine的原型匹配
调用c_f_procpointer（func，cfun）
 
！查找字符串长度
 length = 0 
 do 
 if（str（length + 1）== C_NULL_CHAR）exit 
 length = length + 1 
 end do 
 
 print *，'在Fortran中，得到了string：'，（str（i），i = 1，length），'（'，length，'）。'
 
 print * ，'在Fortran中调用C函数并传递长度'
 clen = length 
 getstring = cfun（clen）
 
结束函数getstring 
  
 
 
结果：
 
 
  $ gcc  - g -Wall -c -o cprogram.o cprogram.c 
 $ gfortran -c fortranroutine.f90 -g -Wall 
 $ gcc -o cprogram cprogram.o fortranroutine.o -lgfortran -g -Wall 
 $ gpc-f103n084- $ ./cprogram 
 ./cprogram 
在Fortran中，得到字符串：IGRF（4）。 
在Fortran中，调用C函数并传递长度
在Fortran中调用的C中：4平方为16！ 
在C中：l = 16 
在Fortran中，得到字符串：IGRF（4）。 
在Fortran中，调用C函数并传递长度
在从Fortran中调用的C中：4立方体为64！ 
在C：l = 64 
  
 
I feel like this should be an easy question, but I can't get it to work.  I have some Fortran code that takes an input like:
      SUBROUTINE TRACE(X,Y,NAME,XX,YY)
      EXTERNAL NAME
      CALL NAME(X,Y,XX,YY)
and I'm trying to pass in a name from C++ in the form:
float x,y,xx,yy;
char * name="IGRF";
trace_(&x,&y,name,&xx,&yy);
It compiles, but I always get segfaults when I try to call the NAME subroutine. A subroutine called IGRF is defined in the file, and I can call the IGRF subroutine directly from C++, but need this TRACE routine. When running in gdb, it says the NAME variable comes through as a pointer to void.

I've tried passing NAME, &NAME, &NAME[0], a char NAME[4] that's stripped of its \0 to perfectly fit the name, and they all come back showing the same void pointer. Does anybody know how to get a function name from C++ into that EXTERNAL variable in Fortran?

Thank you
解决方案
So one advantage of Fortran2003 and later is that C interoperability is defined into the standard; it's a bit of a PITA to use, but once it's done, it's guaranteed to work across platforms and compilers.

So here's cprogram.c, calling a Fortran routine getstring:
#include <stdio.h>

int main(int argc, char **argv) {
    int l;
    char *name="IGRF";

    l = getstring(name);

    printf("In C: l = %d\n",l);

    return 0;
}
and here's fortranroutine.f90:
integer(kind=c_int) function getstring(instr) bind(C,name='getstring') 
    use, intrinsic :: iso_c_binding
    character(kind=c_char), dimension(*), intent(IN) :: instr
    integer :: len
    integer :: i

    len=0
    do
       if (instr(len+1) == C_NULL_CHAR) exit
       len = len + 1
    end do


    print *, 'In Fortran:'
    print *, 'Got string: ', (instr(i),i=1,len)
    getstring = len
end function getstring
The makefile is simple enough:
CC=gcc
FC=gfortran

cprogram: cprogram.o fortranroutine.o
    $(CC) -o cprogram cprogram.o fortranroutine.o -lgfortran

fortranroutine.o: fortranroutine.f90
    $(FC) -c $^

clean:
    rm -f *.o cprogram *~
and running it works, under both gcc/gfortran and icc/ifort:
 In Fortran:
 Got string: IGRF
In C: l = 4
Update:  Oh, I just realized that what you're doing is rather more elaborate than just passing a string; you're essentially trying to pass a function pointer pointing to a C callback routine.    That's a little tricker, because you have to use Fortran interfaces to declare the C routine -- just using extern won't work (and isn't as good as explicit interfaces anyway, as there's no type checking, etc.)   So this should work:

cprogram.c:
#include <stdio.h>

/* fortran routine prototype*/
int getstring(char *name, int (*)(int));

int square(int i) {
    printf("In C called from Fortran:, ");
    printf("%d squared is %d!\n",i,i*i);
    return i*i;
}


int cube(int i) {
    printf("In C called from Fortran:, ");
    printf("%d cubed is %d!\n",i,i*i*i);
    return i*i*i;
}

int main(int argc, char **argv) {
    int l;
    char *name="IGRF";

    l = getstring(name, &square);
    printf("In C: l = %d\n",l);
    l = getstring(name, &cube);
    printf("In C: l = %d\n",l);


    return 0;
}
froutine.f90:
integer(kind=c_int) function getstring(str,func) bind(C,name='getstring')
    use, intrinsic :: iso_c_binding
    implicit none
    character(kind=c_char), dimension(*), intent(in) :: str
    type(c_funptr), value :: func

    integer :: length
    integer :: i

    ! prototype for the C function; take a c_int, return a c_int
    interface
        integer (kind=c_int) function croutine(inint) bind(C)
            use, intrinsic :: iso_c_binding
            implicit none
            integer(kind=c_int), value :: inint
        end function croutine
    end interface
    procedure(croutine), pointer :: cfun

    integer(kind=c_int) :: clen

    ! convert C to fortran procedure pointer,
    ! that matches the prototype called "croutine"
    call c_f_procpointer(func, cfun)

    ! find string length
    length=0
    do
       if (str(length+1) == C_NULL_CHAR) exit
       length = length + 1
    end do

    print *, 'In Fortran, got string: ', (str(i),i=1,length), '(',length,').'

    print *, 'In Fortran, calling C function and passing length'
    clen = length
    getstring = cfun(clen)

end function getstring
And the results:
$ gcc -g -Wall   -c -o cprogram.o cprogram.c
$ gfortran -c fortranroutine.f90 -g -Wall
$ gcc -o cprogram cprogram.o fortranroutine.o -lgfortran -g -Wall
$ gpc-f103n084-$ ./cprogram 
./cprogram 
 In Fortran, got string: IGRF(           4 ).
 In Fortran, calling C function and passing length
In C called from Fortran:, 4 squared is 16!
In C: l = 16
 In Fortran, got string: IGRF(           4 ).
 In Fortran, calling C function and passing length
In C called from Fortran:, 4 cubed is 64!
In C: l = 64


                        
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