如何确定在使用指针数组时指向哪个派生类

Question

我正在编写一个C++程序,它将掷骰子并翻转硬币.我需要使用继承和多态.我正确设置了虚拟功能.在我的基类(aRandomNumberGenerator)中,我有一个虚函数生成.在main()中,我需要一个包含2个基类指针的数组,这些指针指向派生类(aDie和aCoin).当我调用generate()函数时,如何知道数组中指向哪个派生类？

码:

int main()
{
    int n;
    int frequency1 = 0; // count of 1s rolled
    int frequency2 = 0; // count of 2s rolled
    int frequency3 = 0; // count of 3s rolled
    int frequency4 = 0; // count of 4s rolled
    int frequency5 = 0; // count of 5s rolled
    int frequency6 = 0; // count of 6s rolled
    int face; // stores most recently rolled value
    int numHeads = 0; // count of heads
    int numTails = 0; // count of tails
    int side; // stores most recently flipped value

    cout << "Enter a seed number: "; // promp for user to enter a seed number
    cin >> n;
    cout << endl;
    cout << endl;

    aRandomNumberGenerator number(n);
    aDie die(n, n);
    aCoin coin(n, n);

    aRandomNumberGenerator *games[2]; // array of 2 base class pointers

    for(int i=0; i <= 2; i++)  
      games[i]->generate();  


    // output the seed number entered by the user
    cout << "the seed entered is: " << die.inputSeed(n) << endl;
    cout << endl;
    cout << endl;

    // summarize results for 600 flips of a coin
    for(int counter2 = 0; counter2 < 600; counter2++)
    {
        side = generate();

        // determine flip value 0/1 and increment appropriate counter
        switch ( side )
        {
        case 0:
            ++numHeads;
            break;
        case 1:
            ++numTails;
            break;
        default:
            cout << "can only be heads or tails";
        }
    }

    // summarize results for 600 rolls of a die
    for(int counter1 = 0; counter1 < 600; counter1++)
    {
        face = generate();

        // determine roll value 1-6 and increment appropriate counter
        switch ( face )
        {
        case 1:
            ++frequency1;
            break;
        case 2:
            ++frequency2;
            break;
        case 3:
            ++frequency3;
            break;
        case 4:
            ++frequency4;
            break;
        case 5:
            ++frequency5;
            break;
        case 6:
            ++frequency6;
            break;
        default:
            cout << "7 doen't exist on a die!";
        }
    }

    // output results
    cout << "Heads: " << numHeads << endl;
    cout << "Tails: " << numTails << endl;
    cout << "1: " << frequency1 << endl;
    cout << "2: " << frequency2 << endl;
    cout << "3: " << frequency3 << endl;
    cout << "4: " << frequency4 << endl;
    cout << "5: " << frequency5 << endl;
    cout << "6: " << frequency6 << endl;

    cout << endl;
    cout << endl;
    system ("pause");
    return 0;

Answer 1

使用dynamic_cast.

if (Derived1* ptr = dynamic_cast<Derived1*>(basepointer)) {
    // Do something with ptr
}

但是,对于常规方法调用,您无需确定它.只需在基类上调用该方法,它就会被调度到正确的派生类 - 这就是继承的目的.

编辑:对于常规虚拟方法,您只需在基指针上调用它,不知道或不关心Derived是什么,您将获得正确的函数调用.