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我一直在使用 Eigen 的 AutoDiffScalar 并取得了很大的成功，现在我想改用 AutoDiffJacobian 而不是自己做这个。因此，我在学习了 AutoDiffJacobian.h 之后创建了一个学习示例，但出现了问题。

函子：

template <typename Scalar>
struct adFunctor
{
  typedef Eigen::Matrix<Scalar, 3, 1> InputType;
  typedef Eigen::Matrix<Scalar, 2, 1> ValueType;
  typedef Eigen::Matrix<Scalar,
                        ValueType::RowsAtCompileTime,
                        InputType::RowsAtCompileTime> JacobianType;

  enum {
    InputsAtCompileTime = InputType::RowsAtCompileTime,
    ValuesAtCompileTime = ValueType::RowsAtCompileTime
  };

  adFunctor() {}

  size_t inputs() const { return InputsAtCompileTime; }

  void operator() (const InputType &input,
                   ValueType *output) const
  {
    Scalar s1 = Scalar(0), s2 = Scalar(0);

    /* Some operations to test the AD. */
    for (int i = 0; i < 3; i++)
    {
      s1 += log(input(i));
      s2 += sqrt(input(i));
    }

    (*output)(0) = s1;
    (*output)(1) = s2;
  }
};

用法：

Eigen::Matrix<double, 3, 1> in;
in << 1,2,3;
Eigen::Matrix<double, 2, 1> out;
Eigen::AutoDiffJacobian< adFunctor<double> > adjac;
adjac(in, &out);

从这里收到的错误如下：

/usr/include/eigen3/unsupported/Eigen/src/AutoDiff/AutoDiffJacobian.h: In instantiation of ‘void Eigen::AutoDiffJacobian<Functor>::operator()(const InputType&, Eigen::AutoDiffJacobian<Functor>::ValueType*, Eigen::AutoDiffJacobian<Functor>::JacobianType*) const [with Functor = adFunctor<double>; Eigen::AutoDiffJacobian<Functor>::InputType = Eigen::Matrix<double, 3, 1>; Eigen::AutoDiffJacobian<Functor>::ValueType = Eigen::Matrix<double, 2, 1>; Eigen::AutoDiffJacobian<Functor>::JacobianType = Eigen::Matrix<double, 2, 3, 0, 2, 3>]’:
/home/emifre/Git/autodiff-test/src/autodiff_test.cpp:55:17:   required from here
/usr/include/eigen3/unsupported/Eigen/src/AutoDiff/AutoDiffJacobian.h:69:24: error: no matching function for call to ‘Eigen::AutoDiffJacobian<adFunctor<double> >::operator()(Eigen::AutoDiffJacobian<adFunctor<double> >::ActiveInput&, Eigen::AutoDiffJacobian<adFunctor<double> >::ActiveValue*) const’
     Functor::operator()(ax, &av);
     ~~~~~~~~~~~~~~~~~~~^~~~~~~~~
/home/emifre/Git/autodiff-test/src/autodiff_test.cpp:27:8: note: candidate: void adFunctor<Scalar>::operator()(const InputType&, adFunctor<Scalar>::ValueType*) const [with Scalar = double; adFunctor<Scalar>::InputType = Eigen::Matrix<double, 3, 1>; adFunctor<Scalar>::ValueType = Eigen::Matrix<double, 2, 1>]
   void operator() (const InputType &input,
        ^~~~~~~~
/home/emifre/Git/autodiff-test/src/autodiff_test.cpp:27:8: note:   no known conversion for argument 2 from ‘Eigen::AutoDiffJacobian<adFunctor<double> >::ActiveValue* {aka Eigen::Matrix<Eigen::AutoDiffScalar<Eigen::Matrix<double, 3, 1> >, 2, 1, 0, 2, 1>*}’ to ‘adFunctor<double>::ValueType* {aka Eigen::Matrix<double, 2, 1>*}’

从这个错误看来，我在第二次调用 AutoDiffJacobian.h 中的函子时不知何故没有正确的函子类型，但第一次调用它有效。我希望这里有人知道为什么并且可以提供帮助，也许我只是误解了用法。

编辑：显示问题的可编译示例：

#include <Eigen/Dense>
#include <unsupported/Eigen/AutoDiff>

/*
 * Testing differentiation that will produce a Jacobian, using functors and the
 * AutoDiffJacobian helper.
 */

template <typename Scalar>
struct adFunctor
{
  typedef Eigen::Matrix<Scalar, 3, 1> InputType;
  typedef Eigen::Matrix<Scalar, 2, 1> ValueType;
  typedef Eigen::Matrix<Scalar,
                        ValueType::RowsAtCompileTime,
                        InputType::RowsAtCompileTime> JacobianType;

  enum {
    InputsAtCompileTime = InputType::RowsAtCompileTime,
    ValuesAtCompileTime = ValueType::RowsAtCompileTime
  };

  adFunctor() {}

  size_t inputs() const { return InputsAtCompileTime; }

  void operator() (const InputType &input,
                   ValueType *output) const
  {
    Scalar s1 = Scalar(0), s2 = Scalar(0);

    /* Some operations to test the AD. */
    for (int i = 0; i < 3; i++)
    {
      s1 += log(input(i));
      s2 += sqrt(input(i));
    }

    (*output)(0) = s1;
    (*output)(1) = s2;
  }
};



int main(int argc, char *argv[])
{
  Eigen::Matrix<double, 3, 1> in;
  in << 1,2,3;
  Eigen::Matrix<double, 2, 1> out;
  Eigen::AutoDiffJacobian< adFunctor<double> > adjac;
  adjac(in, &out);

  return 0;
}
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1 回答 1

3

好的，经过大量测试后，我让它工作了。只是我误解了编译器直接说出的错误，我错过了运算符本身的模板。

它只需要更改为：

template <typename T1, typename T2>
void operator() (const T1 &input, T2 *output) const

现在它就像一种享受！我希望比我更多的人使用这个。

于 2016-09-11T17:40:57.887 回答