c ++通用模板類算術

Question

假設我們正在處理一個由於某種原因必須進行一些算術運算的類。

像tensor_sum這樣的操作有重載的操作符模板。 這種方法的問題似乎是這樣的：

g++ main.cpp -o main
main.cpp: In instantiation of ‘tensor_sum<T0, T1>::value_type& 

tensor_sum<T0, T1>::operator()(unsigned int) const [with T0 = tensor<int>; T1 = tensor<int>; tensor_sum<T0, T1>::value_type = int; typename T0::value_type = int; typename T1::value_type = int]’:
main.cpp:46:20:   required from here
main.cpp:11:63: error: no match for call to ‘(const tensor<int>) (unsigned int&)’
   value_type & operator () (unsigned int i) const { return t0_(i) + t1_(i); }
                                                            ~~~^~~
main.cpp:32:7: note: candidate: T& tensor<T>::operator()(unsigned int) [with T = int] <near match>
   T & operator () (unsigned int i) { return values_[i]; }
       ^~~~~~~~
main.cpp:32:7: note:   passing ‘const tensor<int>*’ as ‘this’ argument discards qualifiers
main.cpp:11:72: error: no match for call to ‘(const tensor<int>) (unsigned int&)’
   value_type & operator () (unsigned int i) const { return t0_(i) + t1_(i); }
                                                                     ~~~^~~
main.cpp:32:7: note: candidate: T& tensor<T>::operator()(unsigned int) [with T = int] <near match>
   T & operator () (unsigned int i) { return values_[i]; }
       ^~~~~~~~
main.cpp:32:7: note:   passing ‘const tensor<int>*’ as ‘this’ argument discards qualifiers

出於某種原因，我無法訪問該值。 但是我重載了() operator

無論如何這里是代碼：

#include <iostream>
#include <vector>

template<typename T0, typename T1>
struct tensor_sum {
  typedef decltype(typename T0::value_type() + typename T1::value_type()) value_type;

  public:
  tensor_sum(const T0 &t0, const T1 &t1) : t0_(t0), t1_(t1) {}

  value_type & operator () (unsigned int i) const { return t0_(i) + t1_(i); }

  private:
  const T0 &t0_;
  const T1 &t1_;
};

template<typename T0, typename T1>
tensor_sum<T0, T1> operator + (const T0 &t0, const T1 &t1) { return tensor_sum<T0, T1>(t0, t1); }

template<typename T0, typename T1>
tensor_sum<T0, T1> operator + (const T0 &t0, const T1 &t1);

template<typename T>
struct tensor {
  typedef T value_type;

  public:
  tensor(const unsigned int s = 0) : size_(s), values_(std::vector<T>(s)) {}
  tensor(const tensor<T> &t) : size_(t.size_), values_(std::vector<T>(t.values_)) {}

  T & operator () (unsigned int i) { return values_[i]; }
  tensor<T> & operator = (const tensor<T> &t) { return tensor<T>(t); }

  private:
  const unsigned int size_;
  std::vector<T> values_;
};

int main() {
  tensor<int> t0(10);
  tensor<int> t1(10);

  tensor_sum<tensor<int>, tensor<int>> ts = t0 + t1;

  std::cout << ts(2) << std::endl; //Can't access value.. why?

  return 0;
}

活生生的例子

Answer 1

tensor_sum<T0, T1> operator + (const T0 &t0, const T1 &t1);

返回一個 tensor_sum 並且 tensor 沒有帶有 tensor_sum 的 operator=。

所以你的代碼意味着 t0 + t1 返回一個 tensor_sum 並嘗試分配給一個張量，當然會失敗。