我怎样才能使用boost ::绑定没有明确的功能定义持有的boost :: shared_ptr的参考？

Question

我要抱着参考对象，以便它不会在绑定功能删除，但不使用辅助功能。

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) );

}

是否有声明holdReference功能到位的方法吗？像lambda表达式或某事？ （不使用这个讨厌的holdReference功能，有乐趣的功能范围之外声明） 我试了几次，但其中非编译：）

确定，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

Answer 1

您可以寻找“共享所有权”构造，这允许参考计数的内部指针。

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);
}

Answer 2

我有点被你正在尝试做的混淆。

时的乐趣（）应该返回boost::shared_ptr<int>或boost::shared_ptr<Int> ???

我不认为你想创建一个shared_ptr<int>是围绕直接被诠释对象拥有的原始指针共享指针，作为诠释对象将删除_INT时，它超出范围（即使在该示例中它没有“新”吧！）。

您需要拿出一个明确的所有权/责任模式。

也许你可以提供你想达到什么不同的，更加逼真，例如？