在C中使用Fortran代码 [英] Use Fortran-code in C
问题描述
模块集成
隐式none
包含
函数Integrate(func,a,b,intsteps)结果(整数)
接口
实函数func x)
real,intent(in):: x
结束函数func
结束接口
real ::整数,a,b
整数: :intsteps
intent(in):: a,b,intsteps
optional :: intsteps
real :: x,dx
integer :: i,n $如果(存在(intsteps))n = intsteps
dx =(ba )/ n
积分= 0.0_rk
do i = 1,n
x = a +(1.0_rk * i - 0.5_rk)* dx
积分=积分+ func(x)
end do
整数=整数* dx
结束函数
结束模块整合
real(c_float)函数n wrapper_integrate(func,a,b,intsteps)结果(整数)bind(C,name ='integrate')
使用iso_c_binding
使用整合
界面
真正的(c_float)函数func(x)绑定(C)
内部:: iso_c_binding
真实(c_float),intent(in):: x
结束函数func
(c_float):: a,b
整数(c_int),intent(in):: intsteps
可选:: intsteps
if(present (func,a,b)
endif
$ b $积分=积分(func,a,b,intsteps)
else
积分= b $ b结束函数wrapper_integrate
和我的C代码:
#include
#include< math.h>
float sin2(float x){
return sin(x)* sin(x);
float(float(* func)(float),float a,float b,int intsteps);
int main(){
float integral;
integral = integrate(sin2,0。,1.,10000);
printf(%f,integral);
返回0;
}
如果我执行
g ++ -c main.c
gfortran -c integration.f95
g ++ main.o integration.o
我得到
(*)(float),float,float,int)'
有谁知道如何处理?
如果您使用模块 ISO_C_Binding
将函数从C传递给Fortran作为函数指针 C_FUNPTR
。
请参阅此处细节。
您的情况如下:
real c_float)函数wrapper_integrate(func,a,b,intsteps)result(整型)bind(C,name ='integrate')
use iso_c_binding
使用整合
抽象接口
函数iFunc(x)bind(C)
use,intrinsic :: iso_c_binding
real(c_float):: iFunc
real(c_float),intent(in):: x
结束函数iFunc
结束接口
类型(C_FUNPTR),INTENT(IN),VALUE :: func
real(c_float):: a,b
整数(c_int),intent(in):: intsteps
可选:: intsteps
过程(iFunc),指针:: myfunc
调用c_f_procpointer(func,myfunc)
if(present(intsteps))then
integral = Integrate(myfunc,a,b,intsteps)
else
integral = Integrate(myfunc,a,b)
endif
结束函数wrapper_integrate
显然,您的解决方案是更优雅; - )另外,请注意,Fortran通过引用传递变量(除非您指定了 VALUE
属性,您不会T)。因此,您需要相应地更改您的C代码:
#include< stdio.h>
#include< math.h>
float sin2(float * x){
return sin(* x)* sin(* x);
float(float(* func)(float *),float * a,float * b,int * intsteps);
int main(){
float integral;
float a = 0 .;
float b = 1 .;
int intsteps = 10000;
积分=积分(sin2,& a,& b,& intsteps);
printf(%f,integral);
返回0;
}
I try to use a fortran-routine in C, but I doesn't work. I don't know where I made a mistake. Here my Fortran-code including the Integration-Module, which I want to use in C:
module integration
implicit none
contains
function Integrate(func, a,b, intsteps) result(integral)
interface
real function func(x)
real, intent(in) :: x
end function func
end interface
real :: integral, a, b
integer :: intsteps
intent(in) :: a, b, intsteps
optional :: intsteps
real :: x, dx
integer :: i,n
integer, parameter :: rk = kind(x)
n = 1000
if (present(intsteps)) n = intsteps
dx = (b-a)/n
integral = 0.0_rk
do i = 1,n
x = a + (1.0_rk * i - 0.5_rk) * dx
integral = integral + func(x)
end do
integral = integral * dx
end function
end module integration
real(c_float) function wrapper_integrate(func,a,b, intsteps) result(integral) bind(C, name='integrate')
use iso_c_binding
use integration
interface
real(c_float) function func(x) bind(C)
use, intrinsic :: iso_c_binding
real(c_float), intent(in) :: x
end function func
end interface
real(c_float) :: a,b
integer(c_int),intent(in) :: intsteps
optional :: intsteps
if (present(intsteps)) then
integral = Integrate(func,a,b,intsteps)
else
integral = Integrate(func,a,b)
endif
end function wrapper_integrate
and my C-Code:
#include <stdio.h>
#include <math.h>
float sin2(float x) {
return sin(x) * sin(x);
}
float integrate(float(*func)(float), float a, float b, int intsteps);
int main() {
float integral;
integral = integrate(sin2,0.,1.,10000);
printf("%f",integral);
return 0;
}
if I execute
g++ -c main.c
gfortran -c integration.f95
g++ main.o integration.o
I get
undefined reference to `integrate(float (*)(float), float, float, int)'
Does anyone know how to handle this?
If you are using the module ISO_C_Binding
, you can directly passing a function from C to Fortran as a function pointer C_FUNPTR
.
See here for details.
In your case, this would look like:
real(c_float) function wrapper_integrate(func, a, b, intsteps) result(integral) bind(C, name='integrate')
use iso_c_binding
use integration
abstract interface
function iFunc(x) bind(C)
use, intrinsic :: iso_c_binding
real(c_float) :: iFunc
real(c_float), intent(in) :: x
end function iFunc
end interface
type(C_FUNPTR), INTENT(IN), VALUE :: func
real(c_float) :: a,b
integer(c_int),intent(in) :: intsteps
optional :: intsteps
procedure(iFunc),pointer :: myfunc
call c_f_procpointer(func, myfunc)
if (present(intsteps)) then
integral = Integrate(myfunc,a,b,intsteps)
else
integral = Integrate(myfunc,a,b)
endif
end function wrapper_integrate
Obviously, your solution is more elegant ;-)
Also, please note that Fortran passes variables by reference (unless you specify the VALUE
attribute, which you don't). Therefore, you need to change you C code accordingly:
#include <stdio.h>
#include <math.h>
float sin2(float *x) {
return sin(*x) * sin(*x);
}
float integrate(float(*func)(float*), float* a, float* b, int* intsteps);
int main() {
float integral;
float a=0.;
float b=1.;
int intsteps=10000;
integral = integrate(sin2, &a, &b, &intsteps);
printf("%f",integral);
return 0;
}
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