c++ - Why is it impossible to get a pointer to a protected method of the base class? -

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इस सवाल का पहले से ही एक उत्तर है: < / P>

  • 6 जवाब 
      वर्ग ए { सार्वजनिक: ए () {ऑटो टीएमपी = & amp; ए :: एफू;} संरक्षित: शून्य foo () {}}; वर्ग बी: सार्वजनिक ए {सार्वजनिक: बी () {ऑटो टीएमपी = & amp; ए :: एफू;}};   
     कक्षा  ए  कोई समस्या नहीं संकलित करता है क्लास <कोड> बी  एक संकलन त्रुटि उत्पन्न करती है:  
     
     'ए :: एफू': वर्ग 'ए' में घोषित संरक्षित सदस्य तक नहीं पहुंच सकता है   < P> ऐसा क्यों है, तर्क क्या है? क्या इसमें कोई रास्ता नहीं है (यदि मुझे कॉलबैक,  std :: function  आदि के लिए उस सूचक की आवश्यकता है?)   
     
     
     यदि आप  a :: foo  पर कोई संकेतक लेने में सक्षम थे, तो आप इसका उपयोग किसी भी वस्तु पर  foo  करने के लिए कर सकते हैं प्रकार  ए  या प्रकार  सी  से व्युत्पन्न  ए :  
      वर्ग बी: सार्वजनिक ए {सार्वजनिक: रहित एक्सएक्स (ए ए) {ऑटो टीएमपी = & amp; ए :: एफू; । एक * tmp (); } // अवैध}   
     इसके बजाय,  B :: foo  पर एक सूचक लें; यह ठीक है क्योंकि आप इसे केवल  B , अपने खुद के वर्ग की वस्तुओं पर उपयोग कर सकते हैं।

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