Question

我对以下代码有问题(这是一个非常简单的例子,

#include <iostream>

using namespace std;

template<class T> class CBase
{
    public:
        template <class T2> CBase(const T2 &x) : _var(x) {;}
        template <class T2> CBase (const CBase<T2> &x) {_var = x.var();}
        ~CBase() {;}
        T var() const {return _var;}
    protected:
        T _var;
};

template<class T> class CDerived : public CBase<T>
{
    public:
        template <class T2> CDerived(const T2 &x) : CBase<T>(x) {;}
        template <class T2> CDerived (const CBase<T2> &x) : CBase<T>(x) {;}
        ~CDerived() {;}
};

int main()
{
    CBase<double> bd(3);
    CBase<int> bi(bd); // <- No problem
    CDerived<double> dd1(3);
    CDerived<double> dd2(dd1);
    CDerived<int> di(dd1); // <- The problem is here
    return 0;
}

错误如下:

error: cannot convert  const CDerived<double>  to  int  in initialization

如何解决这个问题? (优先修改基类而不是衍生类,如果可能,也不使用虚拟性)

非常感谢你

EDIT : If I replace the concerned line with : CDerived<int> di(CBase<int>(CBase<double>(dd1))); it works but it is not very practical...

看来解决的方法是:

template <class T2> CDerived(const CDerived<T2> &x) : CBase<T>(static_cast<const CBase<T2>&>(x)) {;}

Answer 1

CDerived<int> di(dd1); // <- The problem is here

这引用了 Didd < CDerived 的第一个构建者, 因此 > >D2 被推断为 CDerived< ; brother> >, 被推断为 CDerived< breen> , 被传递给接受 Dcode > 的基底类构建者(这是这种类型的参数 d>dd1 到类模板的价值)。因此,错误不能转换为 CDerived< bruny> 无法转换为 < bcode > < ccode > < ccode > < ccode > < ccode > a/code > 。



如下所示:

CDerived<int> di(dd1); // <- The problem is here
          ^       ^
          |       |
          |       this helps compiler to deduce T2 as double
          |
          this is T of the CDerived as well as of CBase


如果您想要使您的代码工作, 那么做到这一点:


First derive publicly instead of privately.
Add another constructor taking CDerived<T2> as parameter.


所以,你需要这样做:

template<class T> class CDerived : public CBase<T>  //derived publicly
{
    public:
        template <class T2> CDerived(const T2 &x) : CBase<T>(x) {;}

        //add this constructor
        template <class T2> CDerived(const CDerived<T2> &x) : CBase<T>(x.var()) {;}

        template <class T2> CDerived (const CBase<T2> &x) : CBase<T>(x) {;}
        ~CDerived() {;}
};


它现在应该能够工作了:在线演示

Answer 2

尝试在基类中另建一个构造器,该构造器将取自一个通用对象,并使用动态铸造来指定值。

template <class T2> CBase (const Object &x) : _var() {
    try {
        const CBase<T2> &x_casted = dynamic_cast<const CBase<T2> &> (x);
        _var = x_casted.var();
    }
    catch {
        std::cerr << "Object not of type CBase" << std::endl; 
    }
}

注意: 这可能会被视为不怎么风格。动态铸造在运行期间比使用 < code> vitual 和超载更昂贵, 所以考虑重新设定代码。

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