c++ - 我可以退回不同类型的模板化容器吗？

Question

我怎样才能创建一个像这样的函数？我似乎无法正确声明返回类型。

template <typename C, typename T0, typename T1>
typename C<T1> 
convert_container(const C<T0>& container, T1 value) {
  C<T1> new_container;
  // Do some stuff...
  return new_container;
}

std::vector<int> vec0;
const auto& vec1 = convert_container(vec0, float(2.0f)); // Produce a vector of floats

std::list<int> lst0;
const auto& lst1 = convert_container(lst0, float(2.0f)); // Produce a list of floats

Answer 1

正确的做法是使用template模板参数:

• C++11:

  template <template<typename...> class C, typename T0, typename T1>
  C<T1> convert_container(const C<T0>& container, T1 value) {
    C<T1> new_container;
    // Do some stuff...
    return new_container;
  }
  

• C++03(带分配器重新绑定(bind)):

  template <template<typename, typename> class C, typename T0, typename T1, typename Alloc>
  C<T1, typename Alloc::template rebind<T1>::other>
  convert_container(const C<T0, Alloc>& container, T1 value) {
    C<T1, typename Alloc::template rebind<T1>::other> new_container;
    // Do some stuff...
    return new_container;
  }
  

• C++03(无需重新绑定(bind)):

  template <template<typename, typename> class C, typename T0, typename T1, typename Alloc>
  C<T1, std::allocator<T1> > convert_container(const C<T0, Alloc>& container, T1 value) {
    C<T1, std::allocator<T1> > new_container;
    // Do some stuff...
    return new_container;
  }