Question

我已经能够在包含这样的方法的c ++ dll中读取函数：

 extern "C" { __declspec(dllexport) Vector4  resta(Vector4 vector1,Vector4 vector2); }

我得到了一个实现以下功能的dll：

 extern "C"{ __declspec(dllexport) void pinverse(Vector4 skel[20], Vector4 result[20]); }

这个函数采用20个vector4变量的数组，并使用armadillo例程计算矩阵的伪逆，但我没有准确地告诉哪些例程，并且当从c＃i执行函数时，我得到一个全零的矩阵。我还从c ++执行了这个函数，同时将armadillo dll保存在projecto文件夹中，并将它们添加到链接器输入中，我得到的结果是非零的。所以问题是，如何包含应该在c＃代码上执行的armadillo例程？如果dll中的例程没有从库中找到其他例程，为什么我没有看到任何错误消息？

我正在使用的c＃中的语法是：

 [DllImport("C:\\Users\\PALMA\\Documents\\Visual Studio 2013\\Projects\\PruebaTest\\PruebaTest\\PruebaDll.dll", CallingConvention = CallingConvention.Cdecl)]
    public static unsafe extern void pinverse(Vector4[] skel, Vector4[] result);

和Vector4是这样定义的结构：

 public struct Vector4{

   public static bool operator !=(Vector4 vector1, Vector4 vector2);
   public static bool operator ==(Vector4 vector1, Vector4 vector2);

   public float W { get; set; }
   public float X { get; set; }
   public float Y { get; set; }
   public float Z { get; set; }

   public override bool Equals(object obj);
   public bool Equals(Vector4 vector);
   public override int GetHashCode();

}

在c＃中我继续创建一个调用导入函数的类，如下所示：

 public static  Vector4[]  ProbarArmadillo(Vector4[] resultado)
      {
    Vector4[] skel1=new Vector4[20];
    Vector4[] skel2=new Vector4[20];

    skel1[0].X = (float)-0.505401; skel1[0].Y = (float)-0.0968128; skel1[0].Z = (float)2.54701; skel1[0].W = 1;
    skel1[1].X = (float)-0.547074; skel1[1].Y = (float)-0.0348387; skel1[1].Z = (float)2.5876; skel1[1].W = 1;
    skel1[2].X = (float)-0.532013; skel1[2].Y = (float)0.323909; skel1[2].Z = (float)2.60893; skel1[2].W = 1;
    skel1[3].X = (float)-0.468823; skel1[3].Y = (float)0.514958; skel1[3].Z = (float)2.63824; skel1[3].W = 1;
    skel1[4].X = (float)-0.642778; skel1[4].Y = (float)0.221897; skel1[4].Z = (float)2.49321; skel1[4].W = 1;
    skel1[5].X = (float)-0.765644; skel1[5].Y = (float)0.172945; skel1[5].Z = (float)2.27342; skel1[5].W = 1;
    skel1[6].X = (float)-0.793357; skel1[6].Y = (float)0.258069; skel1[6].Z = (float)2.04539; skel1[6].W = 1;
    skel1[7].X = (float)-0.802389; skel1[7].Y = (float)0.291529; skel1[7].Z = (float)2.01768; skel1[7].W = 1;
    skel1[8].X = (float)-0.405216; skel1[8].Y = (float)0.194607; skel1[8].Z = (float)2.75541; skel1[8].W = 1;
    skel1[9].X = (float)-0.297749; skel1[9].Y = (float)0.153036; skel1[9].Z = (float)2.85184; skel1[9].W = 1;
    skel1[10].X = (float)-0.164521; skel1[10].Y = (float)0.228597; skel1[10].Z = (float)2.93223; skel1[10].W = 1;
    skel1[11].X = (float)-0.105644; skel1[11].Y = (float)0.269457; skel1[11].Z = (float)2.98627; skel1[11].W = 1;
    skel1[12].X = (float)-0.551655; skel1[12].Y = (float)-0.15617; skel1[12].Z = (float)2.48071; skel1[12].W = 1;
    skel1[13].X = (float)-0.605082; skel1[13].Y = (float)-0.55405; skel1[13].Z = (float)2.37626; skel1[13].W = 1;
    skel1[14].X = (float)-0.659974; skel1[14].Y = (float)-0.880765; skel1[14].Z = (float)2.35654; skel1[14].W = 1;
    skel1[15].X = (float)-0.597098; skel1[15].Y = (float)-0.914582; skel1[15].Z = (float)2.29637; skel1[15].W = 1;
    skel1[16].X = (float)-0.448527; skel1[16].Y = (float)-0.184062; skel1[16].Z = (float)2.58483; skel1[16].W = 1;
    skel1[17].X = (float)-0.49097; skel1[17].Y = (float)-0.628932; skel1[17].Z = (float)2.57352; skel1[17].W = 1;
    skel1[18].X = (float)-0.515788; skel1[18].Y = (float)-0.962651; skel1[18].Z = (float)2.54876; skel1[18].W = 1;
    skel1[19].X = (float)-0.476598; skel1[19].Y = (float)-1.02878; skel1[19].Z = (float)2.53682; skel1[19].W = 1;

    skel2[0].X = (float)-0.505401; skel2[0].Y = (float)-0.0968128; skel2[0].Z = (float)2.54701; skel2[0].W = 1;
    skel2[1].X = (float)-0.547074; skel2[1].Y = (float)-0.0348387; skel2[1].Z = (float)2.5876; skel2[1].W = 1;
    skel2[2].X = (float)-0.532013; skel2[2].Y = (float)0.323909; skel2[2].Z = (float)2.60893; skel2[2].W = 1;
    skel2[3].X = (float)-0.468823; skel2[3].Y = (float)0.514958; skel2[3].Z = (float)2.63824; skel2[3].W = 1;
    skel2[4].X = (float)-0.642778; skel2[4].Y = (float)0.221897; skel2[4].Z = (float)2.49321; skel2[4].W = 1;
    skel2[5].X = (float)-0.765644; skel2[5].Y = (float)0.172945; skel2[5].Z = (float)2.27342; skel2[5].W = 1;
    skel2[6].X = (float)-0.793357; skel2[6].Y = (float)0.258069; skel2[6].Z = (float)2.04539; skel2[6].W = 1;
    skel2[7].X = (float)-0.802389; skel2[7].Y = (float)0.291529; skel2[7].Z = (float)2.01768; skel2[7].W = 1;
    skel2[8].X = (float)-0.405216; skel2[8].Y = (float)0.194607; skel2[8].Z = (float)2.75541; skel2[8].W = 1;
    skel2[9].X = (float)-0.297749; skel2[9].Y = (float)0.153036; skel2[9].Z = (float)2.85184; skel2[9].W = 1;
    skel2[10].X = (float)-0.164521; skel2[10].Y = (float)0.228597; skel2[10].Z = (float)2.93223; skel2[10].W = 1;
    skel2[11].X = (float)-0.105644; skel2[11].Y = (float)0.269457; skel2[11].Z = (float)2.98627; skel2[11].W = 1;
    skel2[12].X = (float)-0.551655; skel2[12].Y = (float)-0.15617; skel2[12].Z = (float)2.48071; skel2[12].W = 1;
    skel2[13].X = (float)-0.605082; skel2[13].Y = (float)-0.55405; skel2[13].Z = (float)2.37626; skel2[13].W = 1;
    skel2[14].X = (float)-0.659974; skel2[14].Y = (float)-0.880765; skel2[14].Z = (float)2.35654; skel2[14].W = 1;
    skel2[15].X = (float)-0.597098; skel2[15].Y = (float)-0.914582; skel2[15].Z = (float)2.29637; skel2[15].W = 1;
    skel2[16].X = (float)-0.448527; skel2[16].Y = (float)-0.184062; skel2[16].Z = (float)2.58483; skel2[16].W = 1;
    skel2[17].X = (float)-0.49097; skel2[17].Y = (float)-0.628932; skel2[17].Z = (float)2.57352; skel2[17].W = 1;
    skel2[18].X = (float)-0.515788; skel2[18].Y = (float)-0.962651; skel2[18].Z = (float)2.54876; skel2[18].W = 1;
    skel2[19].X = (float)-0.476598; skel2[19].Y = (float)-1.02878; skel2[19].Z = (float)2.53682; skel2[19].W = 1;

  //  pinverse(skel1, Resultado);
    pinverse(skel1, resultado);
    return resultado;
}

最后我像这样使用main中的函数：

        Vector4[] resultado = new Vector4[20];

        resultado= Procesar.ProbarArmadillo(resultado);
        Recorder.SaveArmadillo(resultado);

Recorder是一个将输出放在txt文件上的类，但该文件显示的是一个满为零的矩阵。

Answer 1

我认为你的结构声明有点偏。我不确定编组程序如何处理自动实现的属性。我会避免他们。我有这样的结构：

[StructLayout(LayoutKind.Sequential)]
public struct Vector4
{
    public float W;
    public float X;
    public float Y;
    public float Z;
}

p / invoke应该提供方向属性。我还假设您在随机尝试时添加的unsafe也不需要。

[DllImport(dllname, CallingConvention = CallingConvention.Cdecl)]
public static extern void pinverse(
    [In, MarshalAs(UnmanagedType.LPArray, SizeConst = 20)]
    Vector4[] skel, 
    [Out, MarshalAs(UnmanagedType.LPArray, SizeConst = 20)]
    Vector4[] result
);

Answer 2

我终于开始工作了。我重写了这个c ++函数：

void inversa(Vector4 *skel1, Vector4 *result){
 fmat temp(20, 4);

for (int i = 0; i < 20; i++){
    temp(i, 0) = (skel1 + i)->x; temp(i, 1) = (skel1 + i)->y; temp(i, 2) = (skel1 + i)->z; temp(i, 3) = (skel1 + i)->w;
}

fmat B = pinv(temp);
for (int j = 0; j < 20; j++){
    (result + j)->x = B(0, j); (result + j)->y = B(1, j); (result + j)->z = B(2, j); (result + j)->w = B(3, j);
 }

}

然后在像这样的dll项目中编译它：

namespace pdll
{

extern "C"{ __declspec(dllexport) void mensaje(); }
extern "C"{ __declspec(dllexport) void promedio(Vector4 skel1[20], Vector4 skel2[20], Vector4 prom[20]); }
extern "C"{ __declspec(dllexport) void pinverse(Vector4 skel[20], Vector4 result[20]); }
extern "C"{ __declspec(dllexport) void inversa(Vector4 *skel1, Vector4 *result); }

}

并将其导入c＃，如下所示：

    [DllImport("C:\\Users\\PALMA\\Documents\\Visual Studio 2010\\Projects\\LeerDLL\\PruebaDll.dll", CallingConvention = CallingConvention.Cdecl)]
    public static unsafe extern void inversa(Vector4* skel1, Vector4* result);

然后创建了一种这样的包装函数：

 public static unsafe Vector4[] Inverse(Vector4[] esqueleto, Vector4[] matrizinversa) {

    fixed (Vector4* skeletonpointer = esqueleto, inversepointer = matrizinversa)
    {
        inversa(skeletonpointer,inversepointer);
    }

    return matrizinversa;
}

并像这样使用它：

        Vector4[] resultado = new Vector4[20];

        resultado= Procesar.ProbarArmadillo(resultado);
        Recorder.SaveArmadillo(resultado);

，.txt文件的输出为：

enter image description here

我尝试的唯一一件事就是通过引用将参数传递给函数，从汉斯帕斯特那里得到了一个批评传递价值的想法。

希望这可以帮助任何人尝试使用c＃中的armadillo例程。

从另一个dll中定义的c＃调用c ++ dll中的函数

2 个答案: