c++ - 使用指针数组时如何确定指向哪个派生类

我正在编写一个C++程序，它将掷骰子并掷硬币。我需要使用继承和多态性。我的虚拟功能已正确设置。在我的基类(aRandomNumberGenerator)中，我生成了一个虚函数。在main()中，我需要一个包含2个基类指针的数组，这些指针指向我的派生类(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？

代码:

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;

最佳答案

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，它将被分派(dispatch)到正确的派生类-这就是继承的目的。

编辑:对于常规的虚拟方法，您可以仅在基本指针上调用它，而不知道或不在乎Derived是什么，并且您将获得正确的函数调用。

关于c++ - 使用指针数组时如何确定指向哪个派生类，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/3321593/