虚拟成员函数更改 typeid 的结果 - 为什么？

Question

此代码运行良好并打印“yes”：

#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“no”

#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：
为了让事情更混乱，这工作得很好：

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

那么为什么*pA只被视为class一个对象（当没有virtual关键字时）却仍然可以调用child的work()方法？

Answer 1

这就是 typeid 的工作原理。没有虚拟成员的 class 是 non-polymorphic 类型 并且不携带 run-time 类型信息。因此无法在运行时间确定它的类型。

参见[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 表示类型仅在 compile-time 处确定（对于 A* 因此是 A）。

虚拟成员函数更改 typeid 的结果 - 为什么？

Virtual member function changes result of typeid - why?

c++

inheritance

typeid