在没有显式函数定义的情况下,如何使用 boost::bind 在 boost::shared_ptr 中保存引用?

发布于 2024-08-25 06:22:02 字数 1971 浏览 4 评论 0原文

我想保留对对象的引用,这样它就不会在绑定函数中被删除,但不使用辅助函数。

struct Int
{
   int *_int;
   ~Int(){ delete _int; }
};

void holdReference(boost::shared_ptr<Int>, int*) {} // helper

boost::shared_ptr<int> fun()
{
   boost::shared_ptr<Int> a ( new Int ); 
   // I get 'a' from some please else, and want to convert it
   a->_int = new int;

   return boost::shared<int>( a->_int, boost::bind(&holdReference, a, _1) );

}

有没有办法在适当的位置声明holdReference函数?就像 lambda 表达式或者什么? (不使用这个讨厌的holdReference函数,它必须在fun函数的范围之外声明) 我尝试了几次,但都没有编译:)

好的,这里是更详细的示例:

#include <boost/shared_ptr.hpp>
#include <boost/bind.hpp>

// the case looks more or less like this
// this class is in some dll an I don't want to use this class all over my project
// and also avoid coppying the buffer
class String_that_I_dont_have 
{
    char * _data; // this is initialized in 3rd party, and released by their shared pointer

public:
    char * data() { return _data; }
};


// this function I created just to hold reference to String_that_I_dont_have class 
// so it doesn't get deleted, I want to get rid of this
void holdReferenceTo3rdPartyStringSharedPtr( boost::shared_ptr<String_that_I_dont_have>, char *) {}


// so I want to use shared pointer to char which I use quite often 
boost::shared_ptr<char> convert_function( boost::shared_ptr<String_that_I_dont_have> other) 
// 3rd party is using their own shared pointers, 
// not the boost's ones, but for the sake of the example ...
{
    return boost::shared_ptr<char>( 
        other->data(), 
        boost::bind(
            /* some in place here instead of holdReference... */
            &holdReferenceTo3rdPartyStringSharedPtr   , 
            other, 
            _1
        )
    );
}

int main(int, char*[]) { /* it compiles now */ }

// I'm just looking for more elegant solution, for declaring the function in place

I want to hold reference to object so it doesn't get deleted in bind function, but without using helper function.

struct Int
{
   int *_int;
   ~Int(){ delete _int; }
};

void holdReference(boost::shared_ptr<Int>, int*) {} // helper

boost::shared_ptr<int> fun()
{
   boost::shared_ptr<Int> a ( new Int ); 
   // I get 'a' from some please else, and want to convert it
   a->_int = new int;

   return boost::shared<int>( a->_int, boost::bind(&holdReference, a, _1) );

}

Is there a way to declare holdReference function in place? Like with lambda expressions or sth? (without using this nasty holdReference function, that have to be declared outside the scope of fun function)
I had few tries but non of them compiled :)

Ok, here is more detailed example:

#include <boost/shared_ptr.hpp>
#include <boost/bind.hpp>

// the case looks more or less like this
// this class is in some dll an I don't want to use this class all over my project
// and also avoid coppying the buffer
class String_that_I_dont_have 
{
    char * _data; // this is initialized in 3rd party, and released by their shared pointer

public:
    char * data() { return _data; }
};


// this function I created just to hold reference to String_that_I_dont_have class 
// so it doesn't get deleted, I want to get rid of this
void holdReferenceTo3rdPartyStringSharedPtr( boost::shared_ptr<String_that_I_dont_have>, char *) {}


// so I want to use shared pointer to char which I use quite often 
boost::shared_ptr<char> convert_function( boost::shared_ptr<String_that_I_dont_have> other) 
// 3rd party is using their own shared pointers, 
// not the boost's ones, but for the sake of the example ...
{
    return boost::shared_ptr<char>( 
        other->data(), 
        boost::bind(
            /* some in place here instead of holdReference... */
            &holdReferenceTo3rdPartyStringSharedPtr   , 
            other, 
            _1
        )
    );
}

int main(int, char*[]) { /* it compiles now */ }

// I'm just looking for more elegant solution, for declaring the function in place

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。

评论(2

多情癖 2024-09-01 06:22:14

我对你在这里尝试做的事情有点困惑。

fun() 应该返回 boost::shared_ptr 还是 boost::shared_ptr???

我认为您不想创建一个 shared_ptr 它是围绕 Int 对象直接拥有的原始指针的共享指针,因为 Int 对象每当它删除 _int 时都会删除 _int超出范围(即使在示例中它没有“新”它!)。

您需要提出一个清晰的所有权/责任模型。

也许您可以提供一个不同的、更现实的例子来说明您想要实现的目标?

I am a bit confused by what you are trying to do here.

Is fun() supposed to return boost::shared_ptr<int> or boost::shared_ptr<Int>???

I don't think that you want to create a shared_ptr<int> that is a shared pointer around a raw pointer that is directly owned by the Int object, as the Int object will delete the _int whenever it goes out of scope (even though in the example it did not 'new' it!).

You need to come up with a clear ownership/responsibility model.

Perhaps you can provide a different, more realistic, example of what you are trying to achieve?

彼岸花ソ最美的依靠 2024-09-01 06:22:13

您可能正在寻找“共享所有权”构造函数,这允许对内部指针进行引用计数。

struct Int
{
   int *_int;
   ~Int(){ delete _int; }
};

boost::shared_ptr<int> fun()
{
   boost::shared_ptr<Int> a (new Int);
   a->_int = new int;

   // refcount on the 'a' instance but expose the interior _int pointer
   return boost::shared_ptr<int>(a, a->_int);
}

You may be looking for the "shared ownership" constructor, this allows ref counting an interior pointer.

struct Int
{
   int *_int;
   ~Int(){ delete _int; }
};

boost::shared_ptr<int> fun()
{
   boost::shared_ptr<Int> a (new Int);
   a->_int = new int;

   // refcount on the 'a' instance but expose the interior _int pointer
   return boost::shared_ptr<int>(a, a->_int);
}
~没有更多了~
我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
原文