Question

我想添加一个操作符覆盖来执行赋值 / __ set__s内联。

模板： -

class CBase {
    public :
        static void SetupVmeInterface(CVmeInterface *in);

    protected :
        static CVmeInterface *pVmeInterface;
};


template <class T> class TCVmeAccess : public CBase {
    public:
        TCVmeAccess(int address);

        T get()
        {
            unsigned long temp = pVmeInterface->ReadAddress(Address);
            T ret = *reinterpret_cast<T*>(&temp);
            return ret;
        };

        T *operator->();
        unsigned long asLong();

        bool set(T data)
        {
            unsigned long write_data = *reinterpret_cast<unsigned long*>(&data);
            return pVmeInterface->WriteAddress(Address, write_data);
        };

        // void operator->(T);
        void operator=(T data)
        { set(data); }

    private :
        int Address;
};

将在模板中使用的结构： -

typedef struct
{
    int a: 1; // 0
    int b: 1; // 1
    int c: 1; // 2
    int d: 1; // 3
    int NotUsed : 28; // 31-4
} _HVPSUControl;

代码正文： -

TCVmeAccess<_HVPSUControl> HVPSUControl(constHVPSUControlBlock);
_HVPSUControl hvpsu = HVPSUControl.get(); // Yep, good, but not as nice as...
int a = HVPSUControl2.get().OperationalRequestPort; // yep, also good, but...
int b = HVPSUControl->a; // works, and is all go so far

HVPSUControl.set(hvpsu); // works, but need _HVPSUControl type
HVPSUControl = hvpsu;    // also works, as operator = is used, but still need type

// this line does not work!
// as the = assignment is redirected into a copy of the struct, not the template
HVPSUControl->a = 1; // this line

那么，有没有办法让上面的这一行工作？

编辑：在中，我希望“this line”作为“set”执行，就像在模板类中一样。

编辑： 1.直接在构造模板的结构的成员中直接分配值的。
2.导致该分配通过模板访问器。

所以我不必在作业上这样做： -

// HVPSUControl is predefined and used many times.
_HVPSUControl hvpsu;
hvpsu.a = 1;
HVPSUControl.set(hvpsu);

我想做

HVPSUControl.a = 1; // or 
HVPSUControl->a = 1; // or ?

随着上线工作：

if（HVPSUControl-＆gt; a）

Answer 1

而不是覆盖“ - ＆gt;”和“=”运算符，你可以从模板结构派生。

template <class T> class TCVmeAccess : public CBase, public T {
    public:
        TCVmeAccess(int address);

        T get();
        // T *operator->();
        unsigned long asLong();

        bool set(T);
        // void operator->(T);
        // void operator=(T);

    private :
        int Address;
};

HVPSUControl.a = 1; // and use this for setting a bitfield.

编辑：如果你想使用自定义赋值运算符，你应该在HVPSUControl或它的基类中声明它，如果你有更多这种类似控件的结构。

struct _HVPSUControl
{
    int a: 1; // 0
    int b: 1; // 1
    int c: 1; // 2
    int d: 1; // 3
    int NotUsed : 28; // 31-4
    void operator = (int x);
};

或

struct _HVPSUBase {
    void operator = (int x);
}
struct _HVPSUControl: public _HVPSUBase
{
    int a: 1; // 0
    int b: 1; // 1
    int c: 1; // 2
    int d: 1; // 3
    int NotUsed : 28; // 31-4
};

Answer 2

你不能让这条线以任何其他方式工作。您致电TCVmeAccess::operator->，返回_HVPSUControl*，然后访问_HVPSUControl*::a。

或者您想将整数1分配给完整的_HVPSUControl结构？

template <class T> class TCVmeAccess : public CBase {
public:
    // ...

    void operator=(T data) { set(data); }
    void operator=(int n) { operator=(T(n)); }

    // ...
};


typedef struct
{
    _HVPSUControl(int n) {
       *this = reinterpret_cast<_HVPSUControl>(n);
    }

    int a: 1; // 0
    int b: 1; // 1
    int c: 1; // 2
    int d: 1; // 3
    int NotUsed : 28; // 31-4
} _HVPSUControl;

然后再做

HVPSUControl = 1;

带有重写运算符的模板类

2 个答案: