c++ - 虚拟成员函数更改 typeid 的结果 - 为什么？

Question

此代码工作正常并打印"is":

#include <iostream>
using std::cout;
using std::endl;

class A {
public:
     virtual void talk() {  cout << "person talking" << endl;}
};

class B : public A {
public:
    void work() { cout << "employee working" << endl; }
};

int main() {
    A* pA = new B();

    if (typeid(*pA) == typeid(B))
        cout << "yes" << endl; // this is printed
    else
        cout << "no" << endl;
}

但如果我删除 virtual 关键字，它将打印“否”

#include <iostream>
using std::cout;
using std::endl;

class A {
public:
     void talk() {  cout << "person talking" << endl;}
};

class B : public A {
public:
    void work() { cout << "employee working" << endl; }
};

int main() {
    A* pA = new B();

    if (typeid(*pA) == typeid(B))
        cout << "yes" << endl;
    else
        cout << "no" << endl; // this is printed
}

为什么会这样？
为什么方法的虚拟性会影响对象的类型？
编辑:
为了让事情更困惑，这很好用:

((B*)(pA))->work(); // works fine, prints "employee working"

那么为什么 *pA 只被认为是 A 类对象(当没有 virtual 关键字时)但仍然可以调用 child 的 work() 方法？

Answer 1

就是这样typeid作品。没有虚拟成员的类是非多态类型并且不携带运行时类型信息。因此无法在运行时确定它的类型。
见 [expr.typeid]/p2 :

When typeid is applied to a glvalue expression whose type is a polymorphic class type, the result refers to a std​::​type_­info object representing the type of the most derived object (that is, the dynamic type) to which the glvalue refers.

和 [expr.typeid]/p3 :

When typeid is applied to an expression other than a glvalue of a polymorphic class type, the result refers to a std​::​type_­info object representing the static type of the expression.

静态类型意味着类型是在编译时简单地确定的(对于 A*，因此是 A)。