error C4703: 使用了可能未初始化的本地指针变量“p” - SDL检查已改为否，仍出现同样错误

Question

在学习RTTI的使用中， 用到了一个C++ Primer Plus里的一个例子， 但是出现了上述错误，经过搜索， 全网唯一的解决方法是把配置中的SDL检查改为否。
改过后，代码仍然报相同错误。求解！

#include <iostream>
#include <cstdlib>
#include <ctime>

using std::cout;

class Grand
{
private:
    int hold;
public:
    Grand(int h = 0) :hold(h) {}
    virtual void Speak() const { cout << "I am a grand class\n"; }
    virtual int Value() const { return hold; }
};

class Superb :public Grand
{
public:
    Superb(int h = 0) :Grand(h) {}
    void Speak() { cout << "Superb!\n"; }
    virtual void Say() const { cout << "Superb Value is " << Value() << std::endl; }
};

class Magnificant :public Superb
{
private:
    char ch;
public:
    Magnificant(int h = 0, char c = 'A') :Superb(h), ch(c) {}
    void Speak() const { cout << "I am a magnificant class!\n"; }
    void Say() const
    {
        cout << "character: " << ch << "and the integer is : " << Value() << std::endl;
    }
};
    Grand * GetOne();

    int main()
    {
        std::srand(std::time(0));
        Grand *pg;
        Superb *ps;
        for (int i = 0; i < 5; i++)
        {
            pg = GetOne();
            pg->Speak();
            if (ps = dynamic_cast<Superb*> (pg))
                ps->Say();
        }
        return 0;
    }

    Grand * GetOne()
    {
        Grand* p;
        switch (std::rand() % 3)
        {
        case 0: p = new Grand(std::rand() % 100);
            break;
        case 1: p = new Superb(std::rand() % 100);
            break;
        case 2: p = new Magnificant(std::rand() % 100, 'A' + std::rand() % 26);
            break;
        }
        return p;
    }

Answer 1

改进一下代码就行， Grand* p; 改为 Grand* p = NULL;