从函数返回后丢失 RTTI 信息
给定一个类和子类：
class Event {...}
class Note : public Event {...}
注释被克隆并存储在函数 f() 内的指针中。类型信息保留在指针中，可以通过dynamic_cast恢复：
void f()
{
pEvent = pNote->Clone(); // create a clone of a Note
ASSERT(dynamic_cast<Note*>(pEvent)); // check the pointer, here it works
}
现在，从f()返回后，类型信息丢失：
f();
ASSERT(dynamic_cast<Note*>(pEvent)); // -> "Access violation - no RTTI-data"
VS调试器显示有效的指针值（未更改），但不显示派生类, 除了处于 f() 范围内时之外。
从函数返回时，指针的 RTTI 信息如何丢失？
Given a class and subclass:
class Event {...}
class Note : public Event {...}
A Note is Cloned and stored in a pointer within a function f(). The type-information is preserved in the pointer and can be recovered by dynamic_cast:
void f()
{
pEvent = pNote->Clone(); // create a clone of a Note
ASSERT(dynamic_cast<Note*>(pEvent)); // check the pointer, here it works
}
Now, after returning from f() the type-information is lost:
f();
ASSERT(dynamic_cast<Note*>(pEvent)); // -> "Access violation - no RTTI-data"
The VS-debugger shows a valid pointer-value (unchanged), but not the derived class,
other than while beeing in the f()-scope.
How can the RTTI-info for a pointer be lost when returning from a function?
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有一个析构函数不小心损坏了指针。消除此错误后，RTTI 将按预期工作。
There was a destructor accidently doing harm to the pointer. After removing this error, the RTTI works as expected.