c++ - 嵌套初始化列表的构造函数

Question

是否可以有一个通用构造函数接受任何类型的初始化列表，即使其中有嵌套列表？

假设您有一个类的以下部分模板特化，该类接受其构造函数嵌套的初始化列表:

模板 A类；

template <>
class ClassA<4> {

  typedef std::initializer_list<double> list_type;
  typedef std::initializer_list<list_type> llist_type;
  typedef std::initializer_list<llist_type> lllist_type;
  typedef std::initializer_list<lllist_type> initializer_type;

  size_t n_[4] = {0};
  double* data_;

public:

  ClassA(initializer_type l) {

    assert(l.size() > 0);
    assert(l.begin()->size() > 0);
    assert(l.begin()->begin()->size() > 0);
    assert(l.begin()->begin()->begin()->size() > 0);

    size_t m = n_[0] = l.size();
    size_t n = n_[1] = l.begin()->size();
    size_t o = n_[2] = l.begin()->begin()->size();
    n_[3] = l.begin()->begin()->begin()->size();

    data_ = new double[m*n*o*n_[3]];

    int i=0, j=0, k=0, p=0;
    for (const auto& u : l) {
      assert(u.size() == n_[1]);
      for (const auto& v : u) {
        assert(v.size() == n_[2]);
        for (const auto& x : v) {
          assert(x.size() == n_[3]);
          for (const auto& y : x) {
            data_[i + m*j + m*n*k + m*n*o*p] = y;
            ++p;
          }
          p = 0;
          ++k;
        }
        k = 0;
        ++j;
      }
      j = 0;
      ++i;
    }
  }

  size_t size() const {
    size_t n = 1;
    for (size_t i=0; i<4; ++i)
      n *= n_[i];
    return n;
  }

  friend std::ostream& operator<<(std::ostream& os, const ClassA& a) {
    for (int i=0; i<a.size(); ++i)
      os<<" "<<a.data_[i];
    return os<<endl;
  }

};


int main()
{

  ClassA<4> TT = { {{{1.}, {7.}, {13.}, {19}}, {{2}, {8}, {14}, {20}}, {{3}, {9}, {15}, {21}}}, {{{4.}, {10}, {16}, {22}}, {{5}, {11}, {17}, {23}}, {{6}, {12}, {18}, {24}}} };
  cout<<"TT -> "<<TT<<endl;

  return 0;
}

此代码打印:

TT ->  1 4 2 5 3 6 7 10 8 11 9 12 13 16 14 17 15 18 19 22 20 23 21 24

现在，我正在尝试概括构造函数，这样我就不必为每个维度专门化类模板。问题是当我用类似的东西替换构造函数时:

template <class L>
ClassA(std::initializer_list<L> l) {
  cout<<"generic list constructor"<<endl;
}

clang 编译器失败并出现错误:

error: no matching constructor for initialization of 'ClassA<4>

有人能指出为什么会这样吗？模板匹配不适用于初始值设定项列表，可能是因为这是一个新的 C++ 功能？谢谢大家...

编辑

感谢@JohannesSchaub-litb 和@Daniel Frey 的帮助，我能够制作一个非常通用的构造函数，它采用任何维度的 initializer_list。这是结果代码:

template <int d, typename T>
class ClassA {

  size_t n_[d] = {0};
  T* data_;

  template <int D, typename U>
  struct Initializer_list {

    typedef std::initializer_list<typename Initializer_list<D-1,U>::list_type > list_type;

    Initializer_list(list_type l, ClassA& a, size_t s, size_t idx) {

      a.n_[d-D] = l.size();

      size_t j = 0;
      for (const auto& r : l)
        Initializer_list<D-1, U> pl(r, a, s*l.size(), idx + s*j++);
    }
  };

  template <typename U>
  struct Initializer_list<1,U> {

    typedef std::initializer_list<T> list_type;

    Initializer_list(list_type l, ClassA& a, size_t s, size_t i) {

      a.n_[d-1] = l.size();
      if (!a.data_)
        a.data_ = new T[s*l.size()];

      size_t j = 0;
      for (const auto& r : l)
        a.data_[i + s*j++] = r;
    }
  };

  typedef typename Initializer_list<d,T>::list_type initializer_type;

public:

  // initializer list constructor
  ClassA(initializer_type l) : data_(nullptr) {
    Initializer_list<d, T> r(l, *this, 1, 0);
  }

  size_t size() const {
    size_t n = 1;
    for (size_t i=0; i<4; ++i)
      n *= n_[i];
    return n;
  }

  friend std::ostream& operator<<(std::ostream& os, const ClassA& a) {
    for (int i=0; i<a.size(); ++i)
      os<<" "<<a.data_[i];
    return os<<endl;
  }
};

int main()
{

  ClassA<4, double> TT = { {{{1.}, {7.}, {13.}, {19}}, {{2}, {8}, {14}, {20}}, {{3}, {9}, {15}, {21}}}, {{{4.}, {10}, {16}, {22}}, {{5}, {11}, {17}, {23}}, {{6}, {12}, {18}, {24}}} };
  cout<<"TT -> "<<TT<<endl;

  return 0;
}

当然是打印代码

TT ->  1 4 2 5 3 6 7 10 8 11 9 12 13 16 14 17 15 18 19 22 20 23 21 24

我喜欢这个模板元编程的东西!感谢你们帮助解决这个问题。

啊啊

Answer 1

我相信您真正想做的是自动构建正确的类型

template<int S, typename E>
class make_list_type {
public:
  typedef std::initializer_list<
    typename make_list_type<S-1, E>::type
  > type;
};

template<typename E>
class make_list_type<0, E> {
public:
  typedef E type;
};

template<int S>
class ClassA {
  typedef typename make_list_type<S, double>::type initializer_type;

public:
  ClassA(initializer_type l) 
};

至于为什么你的尝试没有成功，参见Templates don't always guess initializer list types