typeid和dynamic_cast，gcc 3.3 [英] typeid and dynamic_cast, gcc 3.3
问题描述
你好，
我对以下小程序的
感到非常惊讶：
- - 削减----
#include" iostream.h"
等级基础{
** public：
** base（）{}
** virtual * ~base（）{};
** void * printType（base *** obj）;
};
派生类：公共基础{
** public：
** virtual * void * doIt（）
** {
**** printType（this）;
**} **
};
void base :: printType（base * obj）{
** cout *<< *" type * is *" *<< * typeid（obj）.name（）*<< * endl;
** cout *< ;< * dynamic_cast< derived **>（obj）*<< * endl;
}
int main（int argc，char * argv []）{
**派生* myObj;
** myObj.doIt（）;
}
< br $>
---- cut ----
打印：
类型是P4base
0xbfffec00
我的期望是看到派生的对于
类型名称。
即使它被识别为base
，RTTI也不应该能够动态转换
它到一个派生的对象报告
不是。
用gcc 3.3测试3.3.2。
有人能开导我吗？
问候，
安德烈亚斯
Hello,
I had been quite suprised by the
following little program:
---- cut ----
#include "iostream.h"
class base {
**public:
**base(){}
**virtual*~base(){};
**void*printType(base***obj);
};
class derived : public base {
**public:
**virtual*void*doIt()
**{
****printType(this);
**}**
};
void base::printType(base * obj) {
**cout*<<*"type*is*"*<<*typeid(obj).name()*<<*endl ;
**cout*<<*dynamic_cast<derived**>(obj)*<<*endl;
}
int main(int argc, char * argv[]) {
**derived*myObj;
**myObj.doIt();
}
---- cut ----
It prints:
type is P4base
0xbfffec00
My expectation was to see "derived" for
the type name.
And even if it is identified as "base"
the RTTI should not be able to dynamic-cast
it to a derived object which it is reported
not to be.
Tested with gcc 3.3 and 3.3.2.
Somebody able to enlighten me?
Greetings,
Andreas
推荐答案
Andreas Sch。写道：
Andreas Sch. wrote:
你好，
我对以下小程序感到非常惊讶：
---- cut - -
#include" iostream.h"
基类{
public：
base（）{}
虚拟〜 base（）{};
void printType（base * obj）;
};
类派生：public base {
public：
virtual void doIt（）
{
printType（this）;
}
};
void base :: printType（base * obj）{
cout<< 类型是 << typeid（obj）.name（）<< ENDL;
这不是虚函数，所以它将被调用类型为''this''
设置为base，即使它是从一个虚函数中调用的。
（virtual-ness不传递调用序列，它基于每个函数的b / b $ b $。）动态转换
cout<< dynamic_cast< derived *>（obj）<< ENDL;
我没有看到这样做的目的：它只是打印出指针
的值。你有没有忘记输入（）它？
}
int main（int argc，char * argv []）{
派生myObj;
myObj.doIt（）;
}
---- cut ----
Hello,
I had been quite suprised by the
following little program:
---- cut ----
#include "iostream.h"
class base {
public:
base(){}
virtual ~base(){};
void printType(base * obj);
};
class derived : public base {
public:
virtual void doIt()
{
printType(this);
}
};
void base::printType(base * obj) {
cout << "type is " << typeid(obj).name() << endl;
This isn''t a virtual function, so it will be called with type of ''this''
set to "base", even though it was called from a virtual function
(virtual-ness doesn''t pass along the calling sequence, its on a
per-function basis.) Dynamic-casting
cout << dynamic_cast<derived *>(obj) << endl;
I don''t see the purpose of doing this: it just prints out the pointer
value. Did you forget to typeid() it?
}
int main(int argc, char * argv[]) {
derived myObj;
myObj.doIt();
}
---- cut ----
--
http://www.it-is-truth.org/
Andreas Sch。写道：
Andreas Sch. wrote:
你好，
我对以下小程序感到非常惊讶：
---- cut - -
#include" iostream.h"
这个标题不是标准的 - 使用#include< iostream>
类库{
public：
base（）{}
virtual~base （）{};
^^^^^^^^'';''不需要
void printType（base * obj）;
};
类派生：公共基地{
public：
virtual void doIt（）
{
printType（this）;
}
};
void base :: printType（base * obj）{
cout<< 类型是 << typeid（obj）.name（）<< ENDL;
^^^^^^^^^^^^^^^ ...这与
cout< < 类型是 << typeid（（base *）0）.name（）<< endl;
cout<< dynamic_cast< derived *>（obj）<< endl;
}
int main（int argc，char * argv []）{
派生myObj;
myObj.doIt（）;
}
----剪切----
它打印：
类型是P4base
0xbfffec00
我的期望是看到派生的对于
类型名称。
即使它被识别为基础
，RTTI也不应该动态地将其转换为派生的报告的对象
不是。
使用gcc 3.3和3.3.2进行测试。
有人能够启发我吗？
Hello,
I had been quite suprised by the
following little program:
---- cut ----
#include "iostream.h" this header is not standard - use #include <iostream>
class base {
public:
base(){}
virtual ~base(){}; ^^^^^^^^ '';'' not needed
void printType(base * obj);
};
class derived : public base {
public:
virtual void doIt()
{
printType(this);
}
};
void base::printType(base * obj) {
cout << "type is " << typeid(obj).name() << endl;
^^^^^^^^^^^^^^^ ... this is the same as
cout << "type is " << typeid( (base *) 0 ).name() << endl;
cout << dynamic_cast<derived *>(obj) << endl;
}
int main(int argc, char * argv[]) {
derived myObj;
myObj.doIt();
}
---- cut ----
It prints:
type is P4base
0xbfffec00
My expectation was to see "derived" for
the type name.
And even if it is identified as "base"
the RTTI should not be able to dynamic-cast
it to a derived object which it is reported
not to be.
Tested with gcc 3.3 and 3.3.2.
Somebody able to enlighten me?
试试这个：
派生类;
#include< iostream>
使用命名空间std;
class base {
public：
base（）{}
virtual~base（）{}
模板< typename T>
void printType（T * obj）
{
cout<< 类型是 << typeid（obj）.name（）<< endl;
cout<< dynamic_cast< derived *>（obj）<<结束;
}
};
派生类：公共基础{
public：
虚拟无效doIt（）
{
printType（this）;
}
};
int main（int argc，char * argv []）{
派生myObj;
myObj.doIt（）;
}
Try this:
class derived;
#include <iostream>
using namespace std;
class base {
public:
base(){}
virtual ~base(){}
template <typename T>
void printType(T * obj)
{
cout << "type is " << typeid(obj).name() << endl;
cout << dynamic_cast<derived *>(obj) << endl;
}
};
class derived : public base {
public:
virtual void doIt()
{
printType(this);
}
};
int main(int argc, char * argv[]) {
derived myObj;
myObj.doIt();
}
Asfand Yar Qazi写道：
Asfand Yar Qazi wrote:
（。 ..）
void base :: printType（base * obj）{
cout<< 类型是 << typeid（obj）.name（）<< endl;
这不是虚函数，所以即使它是从虚函数中调用的，它也会被调用类型为''this''
设置为base。 >（virtual-ness不会传递调用序列，它基于每个函数。）动态转换
(...)
void base::printType(base * obj) {
cout << "type is " << typeid(obj).name() << endl;
This isn''t a virtual function, so it will be called with type of ''this''
set to "base", even though it was called from a virtual function
(virtual-ness doesn''t pass along the calling sequence, its on a
per-function basis.) Dynamic-casting
让我们看看是否我明白了：typeid没有评估内存中的
类型信息，但取决于obj类型的类型
。指针（而不是指向的指针）。
Lets see if I got it: typeid does not evaluate the
type information in memory but depends on the type
of the "obj" pointer (and not on what it is pointing to).
cout<< dynamic_cast< derived *>（obj）<< endl;
cout << dynamic_cast<derived *>(obj) << endl;
我没有看到这样做的目的：它只是打印出指针
值。你有没有忘记输入（）？
I don''t see the purpose of doing this: it just prints out the pointer
value. Did you forget to typeid() it?
我只是想看看dynamic_cast是否失败以及
是否提供了0x0。我预计它会失败，因为打印出的bid字符串打印了基数字样。
感谢您的回答！
Andreas
I just wanted to see if the dynamic_cast fails and
delivers 0x0. I expected it to fail since the typeid-line
printed "base" and not "derived".
Thanks for your answer!
Andreas
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