template<typename T1>
void printList(const T1 & t1)
{
    for(typename T1::const_iterator it =t1.begin(); it!=t1.end(); ++it)
    {
      cout<<(*it) << endl;
    }
};

如果写得更好，就得重载 operation << 了。

请注意 typename T1::const_iterator it ;一定要加上typename。即使有些编译器让你通过。

不写的话可能有以下错误：

• main.cpp:103: error: expected `;' before ‘int’
• main.cpp:103: error: ‘it’ was not declared in this scope
• main.cpp:91: instantiated from here
• main.cpp:103: error: dependent-name ‘std::map<T,A,std::less<_Key>,std::allocator<std::pair<const _Key, _Tp> > >::const_iterator’ is parsed as a non-type, but instantiation yields a type
• main.cpp:103: note: say ‘typename std::map<T,A,std::less<_Key>,std::allocator<std::pair<const _Key, _Tp> > >::const_iterator’ if a type is mean 

  如下有问题：

  template<class T, class A>
  void ShowMap(const map<T, A>& v)
  {
    for (map<T, A> ::const_iterator ci = v.begin();ci != v.end(); ++ci)
      cout << ci ->first <<": " << ci ->second <<endl;
      cout << endl;
  }

  解释：It accurately doesn't treat map<T, A>::const_iterator as a type because it relies on template parameters, T and A. To make the compiler believe you, you need to use the typename keyword.