push_back与派生类 [英] push_back with derived class

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本文介绍了push_back与派生类的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我想创建一个对象数组,我想使用某个构造函数。

I want to create an array of objects and I want to use a certain constructor.

std::vector<Modul> arrDigOut; 
arrDigOut.push_back(Modul(IDC_CHECK1, this, "GVL.DigOut1", pAddr));

只要 DigOut 派生类。当我得到它并使用类 DigOut 它失败:

This works as long as DigOut is not a derived class. When I derive it and use the class DigOut it fails:

class Modul
{
protected:
    int         id;
    int         nErr;
    void*       plcVar;
    bool        bDigOut;
    PAmsAddr    pAddr;
    ULONG       lHdlVar;
    CButton*    pBt;

public:
    Modul();

    //Modul(int ID, Cbeckhoff_frontendDlg* pCbeckhoff,void* pVar,PAmsAddr pAdr)
    //{
    //    SetID(ID,pCbeckhoff);
    //    plcVar = pVar;
    //    pAddr = pAdr;
    //};

    int         GetID() { return (id); };
    void*       GetplcVar() { return plcVar; };
    void        SetID(int ID, Cbeckhoff_frontendDlg* pCbeckhoff);
    int         InitCheck(Cbeckhoff_frontendDlg* pCbeckhoff);
    CButton*    GetpBt(){return pBt;};
    void        SetButton( Cbeckhoff_frontendDlg* pCbeckhoff);
};

class DigOut : protected Modul
{
public:
    DigOut();
    DigOut(int ID, Cbeckhoff_frontendDlg* pCbeckhoff,void* pVar,PAmsAddr pAdr)
    {
        SetID(ID,pCbeckhoff);
        plcVar = pVar;
        pAddr = pAdr;
    };
};


推荐答案

>。你需要有一个指针的向量或智能指针的基类:

That's because of object slicing. You'll need to have a vector of pointers or smart pointers to the base class:

std::vector<Modul*> arrDigOut; 
arrDigOut.push_back(new Modul(IDC_CHECK1, this,"GVL.DigOut1",pAddr));
arrDigOut.push_back(new DigOut());

如果不使用智能指针,请记住释放内存。

Remember to free the memory if not using smart pointers.

当你有一个 std :: vector< Modul> arrDigOut; ,并插入一个派生类型的对象,它将被切片并成为一个 Modul 。对象的 Modul 部分被复制到向量中,其余部分丢失。

When you have a std::vector<Modul> arrDigOut; and you insert on object of a derived type, it will be sliced and becomes a Modul. The Modul part of the object is copied in the vector, and the rest is lost.

这篇关于push_back与派生类的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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