一个Traits例程

阅读stl源码剖析第三章，写了个traits例程， foo函数可以对iterator的类型做出推断，包括原生指针。

#include <iostream>
using namespace std;
template <typename T>
class MyIterator {
public:
    typedef T value_type;
};

template <typename T>
class Traits {
public:
    typedef typename T::value_type value_type;
};

template <typename T>
class Traits<T *> {
public:
    typedef T value_type;
};

template <typename I>
void foo(I i) {
    //typename means type, not member function
    cout << "typeid = " << typeid(typename Traits<I>::value_type).name() << endl;
}

int main() {
    MyIterator<char> mi;
    MyIterator<string> ms;
    int *pi = NULL;
    foo(mi);
    foo(ms);
    foo(pi);
}