3

我有以下课程：

class MyInteger
{
private:
    __int64 numero;

    static __int64 int64Pow(__int64, __int64);
public:
    // It doesn't matter how these methods are implemented
    friend class MyInteger;
    MyInteger(void);
    MyInteger(const MyInteger&);
    MyInteger(const __int64&);
    ~MyInteger(void);

    static MyInteger const minValue;
    static MyInteger const maxValue;

    MyInteger& operator = (const MyInteger&);
    MyInteger operator + (const MyInteger&) const;
    MyInteger operator - (const MyInteger&) const;
    MyInteger operator * (const MyInteger&) const;
    MyInteger operator / (const MyInteger&) const;
    MyInteger& operator += (const MyInteger&);
    MyInteger& operator -= (const MyInteger&);
    MyInteger& operator *= (const MyInteger&);
    MyInteger& operator /= (const MyInteger&);
    MyInteger operator % (const MyInteger&) const;
    MyInteger& operator %= (const MyInteger&);
    MyInteger& operator ++ ();
    MyInteger operator ++ (int);
    MyInteger& operator -- ();
    MyInteger operator -- (int);
    bool operator == (const MyInteger&) const;
    bool operator != (const MyInteger&) const;
    bool operator > (const MyInteger&) const;
    bool operator < (const MyInteger&) const;
    bool operator >= (const MyInteger&) const;
    bool operator <= (const MyInteger&) const;

    int toStdInt() const
    {
        return (int)numero;
    }
    float toStdFloat() const;
    double toStdDouble() const;
    char toStdChar() const;
    short toStdShortInt() const;
    long toStdLong() const;
    long long toStdLongLong() const;
    unsigned int toStdUInt() const;
    __int64 toStdInt64() const;
    unsigned __int64 toStdUInt64() const;
    unsigned long long int toStdULongLong() const;
    long double toStdULongDouble() const;

    template<class Type>
    Type& operator[](Type* sz)
    {
        return sz[toStdULongLong()];
    }
};

template<class Type>
Type* operator+(const Type* o1, const MyInteger& o2)
{
    return ((o1) + (o2.toStdInt()));
}

我想使用这个类来访问这样的数组元素：

MyInteger myInt(1);
int* intPtr = (int*)malloc(sizeof(int) * N);
intPtr[myInt] = 1;

我以为功能

template<class Type>
Type* operator+(const Type* o1, const MyInteger& o2)
{
    return ((o1) + (o2.toStdInt()));
}

可以解决我的问题，因为正如这篇文章所报告的（C++ 中的数组索引类型）“表达式 E1[E2] 与 *((E1)+(E2)) 相同（根据定义）”，但我收到 C2677 错误（“[”运算符：未找到采用“MyInteger”类型的全局运算符（或没有可接受的转换））

有人可以澄清我这种情况吗？谢谢

4

1 回答 1

3

您可以通过以int类似于以下方式覆盖 MyInteger 类的强制转换来做到这一点：

class MyInteger {
   ...

   operator int() const
   {
       return toStdInt(); /** Your class as an array index (int) */
   }

   ...
}
于 2013-09-12T12:54:17.783 回答