Question

假设我们正在处理一个出于某种原因必须进行算术运算的类。

像tensor_sum这样的操作都有重载的运算符模板。这种方法的问题似乎是这样的：

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

由于某种原因，我无法访问该值。但是我已经使() operator

超载了

反正这里是代码：

#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;
}

Live example

Answer 1

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

返回一个tensor_sum，并且张量没有带tensor_sum的operator =。

所以您的代码意味着t0 + t1返回一个tensor_sum并尝试分配给一个张量，当然会失败。

C ++通用模板化类算法

1 个答案: