这是奇怪的行为吗？ [英] Is this strange behavior?

问题描述

以下是我的代码：


包含< iostream>


类测试{

public：

Test（）{};

void print（）{std :: cout<< "确定]; << std :: endl; };

};


int main（int argc，char ** argv）{

Test * test = new Test （）;

测试r = *测试;


删除测试;

r.print（）; //这一行，这一行

}


我认为评论行会产生错误。

但输出OK通常，这种行为是对的吗？

解决方案

嗨


Bo Yang写道：
< blockquote class =post_quotes>
int main（int argc，char ** argv）{

Test * test = new Test（）;

Test r = *测试;


删除测试;

r.print（）; //这一行，这一行

}


我认为评论行会产生错误。

但输出OK通常，这种行为是对的吗？

当然，你打印一份仍然存在的测试副本。


Markus

Bo Yang< st ****** @ mail.nankai.edu.cnwrites：

int main（int argc，char ** argv）{

测试*测试=新测试（）;

测试r = *测试;


删除测试;

r.print（）; //这一行，这一行

}


我认为评论行会产生错误。

No.

但输出正常，这种行为是对的吗？

是的。 r是一个局部变量（自动），它仍然在该行的范围内（实际上在这个块中）。它不是

对象测试指针的副本，而是测试值的副本到自己的内存中。


干杯，

Rudiger

" Bo Yang" < st ****** @ mail.nankai.edu.cn写信息

新闻：ek ********** @ news.cn99.com ...

以下是我的代码：


包括< iostream>


class测试{

public：

Test（）{};

void print（）{std :: cout<< "确定]; << std :: endl; };

};


int main（int argc，char ** argv）{

Test * test = new Test （）;

测试r = *测试;


删除测试;

r.print（）; //这一行，这一行

}


我认为评论行会产生错误。

为什么这么认为？

但它输出正常，这是行为对吗？

是的，它是。


-Mike

Following is my code:

include <iostream>

class Test{
public:
Test(){};
void print(){ std::cout << "OK" << std::endl ; };
};

int main( int argc , char ** argv ){
Test * test = new Test();
Test r = *test ;

delete test ;
r.print(); //this line , this line
}

I think the comment line will produce an error.
But it output OK normally, is this behavior right?

解决方案

Hi

Bo Yang wrote:

int main( int argc , char ** argv ){
Test * test = new Test();
Test r = *test ;

delete test ;
r.print(); //this line , this line
}

I think the comment line will produce an error.
But it output OK normally, is this behavior right?

Absolutely, you call print on a copy of test that is still existent.

Markus

Bo Yang <st******@mail.nankai.edu.cnwrites:

int main( int argc , char ** argv ){
Test * test = new Test();
Test r = *test ;

delete test ;
r.print(); //this line , this line
}

I think the comment line will produce an error.

No.

But it output OK normally, is this behavior right?

Yes. r is a local variable (automatic), which is still in scope at
that line (actually in this block). It''s not a copy of the pointer of
object test, but a copy of the value of test into its own memory.

Cheers,
Rudiger

"Bo Yang" <st******@mail.nankai.edu.cnwrote in message
news:ek**********@news.cn99.com...

Following is my code:

include <iostream>

class Test{
public:
Test(){};
void print(){ std::cout << "OK" << std::endl ; };
};

int main( int argc , char ** argv ){
Test * test = new Test();
Test r = *test ;

delete test ;
r.print(); //this line , this line
}

I think the comment line will produce an error.

Why do you think so?

But it output OK normally, is this behavior right?

Yes, it is.

-Mike

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