为什么这个取模板参数的函数没有编译

我试图用我在上一个问题中得到的优秀答案来理解这个问题，但我不明白为什么这个代码不能编译。

#include <iostream>
#include <ostream>
#include <string>
#include <cstdlib>
#include <vector>
#include <memory>
template <typename T,
          template <typename ELEM,
                    typename = std::allocator<ELEM> >
                    class CONT = std::deque>
class Stack {
public:
  typedef typename CONT<T>::size_type size_type;
  private:
    CONT<T> elems;         // elements
  public:
    void push(T const&);   // push element
    void pop();            // pop element
    T top() const;         // return top element
    bool empty() const {   // return whether the stack is empty
        return elems.empty();
    }
       size_type size();
};
template <typename T, template <typename,typename> class CONT>
void Stack<T,CONT>::push (T const& elem)
{
    elems.push_back(elem);    // append copy of passed elem
}
template<typename T, template <typename,typename> class CONT>
void Stack<T,CONT>::pop ()
{
    if (elems.empty()) {
        throw std::out_of_range("Stack<>::pop(): empty stack");
    }
    elems.pop_back();         // remove last element
}
template <typename T, template <typename,typename> class CONT>
T Stack<T,CONT>::top () const
{
    if (elems.empty()) {
        throw std::out_of_range("Stack<>::top(): empty stack");
    }
    return elems.back();      // return copy of last element
}

template<typename T,
template<typename, typename= std::allocator<T> > class CONT>
typename Stack<T, CONT>::size_type Stack<T, CONT>::size()
{
  return elems.size();
}
template <typename T,
          template <typename ELEM, typename = std::allocator<ELEM> > class CONT>
static inline void Pout(const CONT<T>& container)
{
  typedef typename CONT<T >::size_type size_type;
  size_type idx = 0;
  size_type sz = CONT<T>::size();
  CONT<T> temp = container;
  std::cout << '[';
  while (idx < sz)
  {
    std::cout << temp.top();
    temp.pop();
    idx++;
    if (idx == sz) break;
    std::cout << ", ";
  }
  std::cout << "]";
}

int main()
{
  try {
    Stack<int,   std::vector > vStack;
    //...
    vStack.push(42);
    vStack.push(7);
    Pout<Stack<int,   std::vector > >(vStack);
  }
  catch (std::exception const& ex) {
    std::cerr << "Exception: " << ex.what() << std::endl;
  }
}

我使用g++4.3.4:得到编译器错误

stack8test.cpp: In function int main():
stack8test.cpp:28: error: no matching function for call to Pout(Stack<int, std::vector>&)
make: *** [stack8test.o] Error 1

我会感谢你的帮助。

我做了一些更改，注释掉了函数调用Pout>（vStack），并将vStack的内容打印到std:：cout。这是新代码。它按预期编译和工作：

#include <iostream>
#include <ostream>
#include <string>
#include <cstdlib>
#include <vector>
#include <memory>
#include <deque>
#include <stdexcept>
template <typename T,
          template <typename ELEM,
                    typename = std::allocator<ELEM> >
                    class CONT = std::deque>
class Stack {
public:
  typedef typename CONT<T>::size_type size_type;
  typedef typename CONT<T>::value_type value_type;
  private:
    CONT<T> elems;         // elements
  public:
    void push(T const&);   // push element
    void pop();            // pop element
    T top() const;         // return top element
    bool empty() const {   // return whether the stack is empty
        return elems.empty();
    }
    size_type size();
    void Pout();
};
template <typename T, template <typename,typename> class CONT>
void Stack<T,CONT>::push (T const& elem)
{
    elems.push_back(elem);    // append copy of passed elem
}
template<typename T, template <typename,typename> class CONT>
void Stack<T,CONT>::pop ()
{
    if (elems.empty()) {
        throw std::out_of_range("Stack<>::pop(): empty stack");
    }
    elems.pop_back();         // remove last element
}
template <typename T, template <typename,typename> class CONT>
T Stack<T,CONT>::top () const
{
    if (elems.empty()) {
        throw std::out_of_range("Stack<>::top(): empty stack");
    }
    return elems.back();      // return copy of last element
}

template<typename T,
template<typename, typename= std::allocator<T> > class CONT>
typename Stack<T, CONT>::size_type Stack<T, CONT>::size()
{
  return elems.size();
}

template <typename T,
template<typename, typename = std::allocator<T> > class CONT>
void Stack<T, CONT>::Pout()
{
  size_type idx = 0;
  size_type sz = size();
  CONT<T>  temp(elems); // make a temp copy of the underlying container and print the temp, since printing is destructive. Note that the underlying CONT must already support copy constructor.
  std::cout << std::endl << '[';
  while (idx < sz)
  {
    std::cout << temp.back();
    temp.pop_back();
    idx++;
    if (idx == sz) break;
    std::cout << ", ";
  }
  std::cout << "]" << std::endl;;
}
template <typename T,
          template <typename ELEM, typename = std::allocator<ELEM> > class CONT>
void Pout(const CONT<T>& container)
{
  typedef typename CONT<T >::size_type size_type;
  size_type idx = 0;
  size_type sz = CONT<T>::size();
  CONT<T> temp = container;
  std::cout << '[';
  while (idx < sz)
  {
    std::cout << temp.top();
    temp.pop();
    idx++;
    if (idx == sz) break;
    std::cout << ", ";
  }
  std::cout << "]";
}

int main()
{
  try {
    Stack<int,   std::vector > vStack;
    //...
    vStack.push(42);
    vStack.push(7);
//    Pout<Stack<int,   std::vector > >(vStack);
    vStack.Pout();
    std::cout << "vStack = [" << vStack.top(); vStack.pop();
    std::cout << ", " << vStack.top() << "]" << std::endl; vStack.pop();
  }
  catch (std::exception const& ex) {
    std::cerr << "Exception: " << ex.what() << std::endl;
  }
}

编译器输出为：

g++ -O2 -g -Wall -c -o stack8test stack8test.cpp

运行程序，我得到：

./stack8test 
vStack = [7, 42]

我真正的问题是为函数模板Pout找到正确的语法来打印类模板Stack的内容。为了澄清，我的意图是将类模板Stack实现为T类型元素的容器。我想根据std:：vector或std:：deque，甚至std:：list类型的std容器来实现Stack。

我称Pout为：

Pout<int, Stack<int, std::vector>>(vStack);

但编译器仍然说：

stack8test.cpp: In function int main():
stack8test.cpp:137: error: no matching function for call to Pout(Stack<int, std::vector>&)
make: *** [stack8test.o] Error 1

更新：为了解决我最初的全局（即非成员）函数问题，将Stack的内容打印到std:：cout，我将Pout（）定义为成员函数。它按预期编译和工作。为全局函数找到正确的语法仍然是一个挑战。