情况源于我的之前一片博客《C++ 智能指针》，在我写demo代码的时候。
向前申明了class Phone, 然后再U_ptr类析构函数中delete Phone的指针。
出现warning C4150: 删除指向不完整“XXX”类型的指针；没有调用析构函数
这个waring会导致内存泄露。前向申明的类的析构函数没有被调用
出现warning的代码如下：
#include <iostream>
using namespace std;
class ConnectManager;
class CommManager;
class Phone;
class U_ptr
{
friend class ConnectManager;
friend class CommManager;
public:
U_ptr(Phone *pH): use(0), pPhone(pH) { }
~U_ptr() { delete pPhone; }
private:
Phone *pPhone;
size_t use;
};
class Phone
{
public:
Phone() { cout << "Hello World!" << endl; }
~Phone() { cout << "Bye!" << endl; }
};
class ConnectManager
{
public:
ConnectManager(U_ptr *p): pUptr(p)
{
++pUptr->use;
cout << "ConnectManager: Hi I get the Phone" << endl;
cout << pUptr->use << " users" << endl;
}
~ConnectManager()
{
cout << "ConnectManaer: Can I delete the Phone?" << endl;
if (--pUptr->use == 0)
{
cout << "Yes, You can" << endl;
delete pUptr;
}
else
{
cout << "No, You can't, The Phone is in use" << endl;
cout << pUptr->use << " users" << endl;
}
}
private:
U_ptr *pUptr;
};
class CommManager
{
public:
CommManager(U_ptr *p): pUptr(p)
{
++pUptr->use;
cout << "CommManager: Hi I get the Phone" << endl;
cout << pUptr->use << " users" << endl;
}
~CommManager()
{
cout << "CommManager: Can I delete the Phone" << endl;
if (--pUptr->use == 0)
{
cout << "Yes, You can" << endl;
}
else
{
cout << "No, You can't. The Phone is in use" << endl;
cout << pUptr->use << " users" << endl;
}
}
private:
U_ptr *pUptr;
};
int main(void)
{
Phone *symbian = new Phone();
U_ptr *pU = new U_ptr(symbian);
ConnectManager connManager = ConnectManager(pU);
CommManager commManager = CommManager(pU);
}
出现原因：
class Phone;这种方式向前申明，其后面的类只能申明其指针，前向申明以后的类无法看到其类实体。
所以，delete的时候，Phone的析构函数对后面的类是透明不可见的，除非使用头文件包含。