模板函数中使用未知类型的容器--如何遍历一个未知容器

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,std::allocator > >::const_iterator’ is parsed as a non-type, but instantiation yields a type

main.cpp:103: note: say ‘typename std::map,std::allocator > >::const_iterator’ if a type is mean 

如下有问题：

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

解释：It accurately doesn't treat map::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.