Question

我试图找到列表中对象的出现次数：

class Complex{
  double re, im; 
 public:
   Complex (double r, double i):re(r), im(i){}
   Complex (){re = im = 0;}

   friend bool    operator == (Complex, Complex); 
};

bool operator == (Complex  a,  Complex b){
 return a.re == b.re and a.im == b.im;
}

template <class ContainerType, class ElementType>
int const count (ContainerType const & container, ElementType const & element){
  int count = 0;
  typename ContainerType::const_iterator i = std::find (container.begin(), container.end(), element);

  while (i != container.end()){
   ++count;
   i = std::find (i + 1, container.end(), element);
  }
 return count;
}

int main(){
 std::list <Complex> lc;
 lc.push_front (Complex (1.2, 3.4));

 std::cout << count (std::string("abhi"), 'a') << '\n';
 std::cout << count (lc, Complex (1.2, 3.4)) << '\n';

 return 0;
}

我用g ++ 4.5得到了这个错误：

templatizedcharOccurences.c++: In function ‘const int count(const ContainerType&, const ElementType&) [with ContainerType = std::list<Complex>, ElementType = Complex]’:
templatizedcharOccurences.c++:51:44:   instantiated from here
templatizedcharOccurences.c++:41:4: error: no match for ‘operator+’ in ‘i + 1’
templatizedcharOccurences.c++:22:9: note: candidate is: Complex operator+(Complex, Complex)

为什么抱怨i+1？显然，我不是迭代器（指针）而不是复杂对象吗？

Answer 1

您可以通过以下两种方式之一简化代码：

使用std::count或其中一个变体进行计数。所以：

return std::count(container.begin(), container.end(), element);

或者只是像你已经做的那样使用一个简单的for循环，但是从头到尾迭代它并自己进行计数，如下所示：

int count = 0;
for (ContainerType::const_iterator it = container.begin(); it != container.end(); ++it)
{
    if (*it == element) ++count;
}
return count;

由于你已经在使用部分STL，我建议使用第一种方法。

Answer 2

似乎Iterator没有重载operator +（int），请尝试：

class Complex{
  double re, im; 
 public:
   Complex (double r, double i):re(r), im(i){}
   Complex (){re = im = 0;}

   friend bool    operator == (Complex, Complex); 
};

bool operator == (Complex  a,  Complex b){
 return a.re == b.re and a.im == b.im;
}

template <class ContainerType, class ElementType>
int const count (ContainerType const & container, ElementType const & element){
  int count = 0;
  typename ContainerType::const_iterator i = std::find (container.begin(), container.end(), element);

  while (i != container.end()){
   ++count;
   i = std::find (++i, container.end(), element);
  }
 return count;
}

int main(){
 std::list <Complex> lc;
 lc.push_front (Complex (1.2, 3.4));

 std::cout << count (std::string("abhi"), 'a') << '\n';
 std::cout << count (lc, Complex (1.2, 3.4)) << '\n';

 return 0;
}

我希望这会奏效。

Answer 3

它是一个没有随机访问操作的列表迭代器。你想做的事：

++i;
i = std::find (i, container.end(), element);

计算列表中对象的出现次数

3 个答案: