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Can a local variable s memory be accessed outside its scope?

考虑以下简单的C++代码:

#include <iostream>

struct Test
{
    Test( int i )
        :   ref( i ),
            ptr( &i ) {}

    int &ref;
    int *ptr;
};

int main()
{
    Test t( 5 );

    std::cout << t.ref << std::endl;
    std::cout << *t.ptr << std::endl;

    return 0;
}

班级测试储存着一个点子,是当地变量的指代器,即生活在 st锁上。我假定,在从测试建筑商返回后,这片人就会被毁。但显然情况并非如此。由于这是该方案的产出:

5
134513968

接触点员的结果是我所期望的:一种随时间变化的随机价值。但是,参考渠道总是在5个方面——正如当地变量仍然存在一样。

谁能向我解释这种行为? I m在64bitlav使用g++(转化4.6.3)。

Regards, enuhtac

Answer 1

http://loungecpp.wikidot.com/faq#toc12” rel=“nofollow noretinger”>un Defin conduct。它只印制了5本;它本可以印刷4390843或hello World,坠毁了你的计算机,燃起了大气,或用你的信用卡购买了Pizza。

此外,由于编辑在产生集会时能够做一些工作,与你的法典相容,因此试图对所发生的事情进行猜测,实际上没有任何意义。这种做法是徒劳无益的(但如果你真的想要知道所有轨道来自哪里,哪里是哪里,那就只是ug。

Answer 2

When constructing t you get the memory address of where i is stored during construction. When you call std::cout << t.ref there has been no reason for that memory location to be used for something else, so the value is still there.

However it s nothing you can trust. It s undefined behaviour to access that memory location after i goes out of scope.

Answer 3

我先仔细检查了它,但仅仅因为你跳下了 st头,并不意味着内容被压倒或改变,该地址的价值保持不变。之后添加一些方法,应当改变。

Answer 4

没有保证立即删除当地变量。它可以保持记忆。但是,这是不明确的行为,它具有危险。

Answer 5

答案是。

You rent a hotel room. You put a book in the top drawer of the bedside table and go to sleep. You check out the next morning, but "forget" to give back your key. You steal the key!

A week later, you return to the hotel, do not check in, sneak into your old room with your stolen key, and look in the drawer. Your book is still there. Astonishing!

How can that be? Isn t the contents of a hotel room drawer inaccessible if you haven t rented the room?

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