c++ 運行期類型識別

有了前面3篇文章的基礎，再看這篇文章就很容易了

這是Loki裏的類型識別的測試，分別測試普通類型，指針類型和類成員指針類型。

下面是測試代碼，測試環境是gcc 4.6.3

NullType.h

#ifndef _NULLTYPE_INC_
#define _NULLTYPE_INC_

class NullType;

#endif

PointerTraits.h

#ifndef _POINTERTRAITS_INC
#define _POINTERTRAITS_INC

#include "NullType.h"

template <typename T>
class TypeTraits
{
private:
    template <class U> struct PointerTraits
    {
        enum {result = false};
        typedef NullType PointeeType;
    };

    template <class U> struct PointerTraits<U*>
    {
        enum {result = true};
        typedef U PointeeType;
    };

    template <class U> struct PToMTraits
    {
        enum {result = false};
    };

    template <class U, class V> struct PToMTraits<U V::*>
    {
        enum {result = true};
    };

public:
    enum {isPointer = PointerTraits<T>::result};
    typedef typename PointerTraits<T>::PointeeType PointeeType;

    enum {isMemberPointer = PToMTraits<T>::result};
};


#endif

main.cpp

#include <iostream>
#include <vector>
using namespace std;

#include "PointerTraits.h"

class T
{
public:
    int a;
};

int main(int argc, char *argv[])
{
    bool iterIsPtr = TypeTraits<vector<int>::iterator>::isPointer;
    cout<<"vector<int>::iterator is "<<(iterIsPtr ? "pointer": "type")<<"\n";

    iterIsPtr = TypeTraits<int*>::isPointer;
    cout<<"int* is "<<(iterIsPtr ? "pointer": "type")<<"\n";

    iterIsPtr = TypeTraits<int*>::isMemberPointer;
    cout<<"int* is member pointer ("<<(iterIsPtr ? "yes": "no")<<")\n";

    /*
     * int T::* 是一個指向類T的int的指針。
     * 如：int T::* c = &T::a;
     */

    //int T::* c = &T::a;
    iterIsPtr = TypeTraits<int T::*>::isMemberPointer;
    cout<<"int* is member pointer ("<<(iterIsPtr ? "yes": "no")<<")\n";

    return 0;
}

編譯：g++ main.cpp

運行：./a.out

輸出：

vector<int>::iterator is type
int* is pointer
int* is member pointer (no)
int T::* is member pointer (yes)