[英]How do you pass Fortran's complex type to C#?
假设我有以下 Fortran 代码
subroutine COMPLEX_PASSING(r, i, c)
!DEC$ ATTRIBUTES DLLEXPORT::COMPLEX_PASSING
REAL*8 :: r, i
COMPLEX*8 :: c
c = cmplx((r * 2), (i * 2))
return
end
Fortran 代码编译用
gfortran -c complex_passing.f90
gfortran -fPIC -shared -o complex_passing.dll complex_passing.o
我如何在 C# 中调用这个子程序? 我尝试了以下代码:
using System;
using System.Runtime.InteropServices;
namespace FortranCalling {
class Program {
static void main(string[] args) {
double real = 4;
double imaginary = 10;
COMPLEX c = new COMPLEX();
complex_passing( ref real, ref imaginary, ref c);
Console.WriteLine("Real: {0}\nImaginary: {1}", c.real, c.imaginary);
Console.ReadLine();
}
[StructLayout(LayoutKind.Sequential)]
struct COMPLEX {
public double real;
public double imaginary;
}
[DllImport("complex_passing.dll", EntryPoint = "complex_passing_", CallingConvention = CallingConvention.Cdecl)]
static extern void complex_passing(ref double r, ref double i, ref COMPLEX c);
}
}
收效甚微 - 我的 COMPLEX 结构似乎正在返回垃圾数据:
Real: 134217760.5
Imaginary: 0
当我期望实部是 8 而虚部是 20 时。
gfortran 将非标准COMPLEX*8
视为大小为 8 字节的复数,实部和虚部各 4 字节。 相反,您需要一个 16 字节的复数,实部和虚部各有 8 个字节( COMPLEX*16
),或者您应该相应地更改 C# 端。
在 gfortran 下可以看到以下效果:
complex*8 :: c8 = (8d0, 20d0)
complex*16 :: c16 = 0
c16%re = TRANSFER(c8,c16)
print*, c8, c16
end
当然,您根本不应该使用complex*
。 使用complex(kind=..)
可以看到参数不匹配。
考虑以下“Fortran”源:
subroutine s(r, i, c)
real(kind(0d0)) :: r, i
complex(kind(0e0)) :: c
c = cmplx((r*2),(i*2))
end subroutine s
interface ! Interface block required to lie to some versions of gfortran
subroutine s(r, i, c)
real(kind(0d0)) :: r, i
complex(kind(0d0)) :: c
end subroutine s
end interface
complex(kind(0d0)) c
call s(4d0, 10d0, c)
print*, c%re
end
并将其与 Fortran 源进行比较:
subroutine s(r, i, c)
real(kind(0d0)) :: r, i
complex(kind(0d0)) :: c
c = cmplx((r*2),(i*2))
end subroutine s
complex(kind(0d0)) c
call s(4d0, 10d0, c)
print*, c%re
end
此外,除了使用kind(0d0)
等之外,还有各种 C 互操作性常量和iso_fortran_env
的存储大小常量。
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