我的虚函数不适用于C++

Question

我已经从我的真实代码编辑了这个,所以它更容易理解.

基类:

class MWTypes
{
public:
    virtual long get() { return (0); }
};

派生类:(还有其他类,如char,double等等......)

class TypeLong : public MWTypes
{
public:
    TypeLong(long& ref) : m_long(ref) {}
    ~TypeLong();

    long get() { return m_long; }
private:
    long& m_long;
};

和存储类:

class RowSet
{
public:
    void addElememnt(MWTypes elem);
    MWTypes getElement();

    std::vector<MWTypes> getVector() { return m_row; }

private:
    std::vector<MWTypes> m_row; 
};

怎么称呼:

for (i = 0; i < NumCols; i++) // NumCols is 3 on this instance
{
    switch(CTypeArray[i]) // this is an int which identifies the type
{  
    case SQL_INTEGER:
    {
    long _long = 0;

    TypeLong longObj(_long);
    MWTypes *ptr = &longObj;

            // some SQL code goes here that changes the value of _long, 
            // there is no need to include it, so this will do.
    _long++;

            // I now want to save the data in a vector to be returned to the user.
    rowSet.addElememnt(*ptr);   

///////////////////////////////////////////////
// some code happens here that is irrelevant //
///////////////////////////////////////////////

// I now want to return the typr I have saved in the vector, 
// I THINK I am doing this right?
    MWTypes returned = rowSet.getElement();

    // lastly I want to get the value in the returned type
    long foo = returned.get();

///////////////////////////////////////////////
// some code happens here that is irrelevant //
///////////////////////////////////////////////

我想我在这里是正确的.'foo'的值始终为0.我感觉这可能是我在向量中存储的方式,或者它可能是基本虚函数,因为它总是返回0.

如果我删除基类中的返回,我会收到LNK2001错误.

Answer 1

 MWTypes returned = rowSet.getElement();

 // lastly I want to get the value in the returned type
 long foo = returned.get();

应该

 MWTypes* returned = &rowSet.getElement();

 // lastly I want to get the value in the returned type
 long foo = returned->get();

要么

 MWTypes& returned = rowSet.getElement(); // actually illegal, but MSVC will let you do

 // lastly I want to get the value in the returned type
 long foo = returned.get();

实际上,必须通过指针或引用进行多态调用.

编辑:这不是你唯一的问题.向量存储对象(而不是指针)的事实将切割对象并破坏其类型信息.

有关其他信息,请参阅此常见问题解答条目,以帮助您解决问题并了解如何调用虚函数.