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技術社區[雲棲]
warning C4150: 刪除指向不完整“XXX”類型的指針；沒有調用析構函數
情況源於我的之前一片博客《C++ 智能指針》，在我寫demo代碼的時候。
向前申明了class Phone, 然後再U_ptr類析構函數中delete Phone的指針。
出現warning C4150: 刪除指向不完整“XXX”類型的指針；沒有調用析構函數
這個waring會導致內存泄露。前向申明的類的析構函數沒有被調用
出現warning的代碼如下：
#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); }
出現原因：
class Phone;這種方式向前申明，其後麵的類隻能申明其指針，前向申明以後的類無法看到其類實體。
所以，delete的時候，Phone的析構函數對後麵的類是透明不可見的，除非使用頭文件包含。
最後更新：2017-04-02 06:51:38