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मेरे पास दो पूर्णांक a, b और इस तरह एक पाश है: 

  के लिए (int I = 0; i & lt; 8; ++ i) {if ((बी एंड (0x01 & lt; & lt; i))! = 0x00) {// c ^ = (a & lt; & lt; i); }}   
 ए और बी चार एरे में संग्रहीत होने पर मैं यह कैसे कर सकता हूँ?  
  अहस्ताक्षरित चार परीक्षण (अहस्ताक्षरित चार * एक [], अहस्ताक्षरित Char * b []) {for (int i = 0; i & lt; 8; ++ i) {if ((b [i] & amp; (0x01 & lt; & lt; i)) = 0x00) {shiftleft (ए) ; सी [आई] ^ = ए; चूंकि ए और बी अब एरेज़ हैं, आपको पहले इंडेक्स होना चाहिए और तुल्यता के साथ काम करना चाहते हैं इसलिए, उदाहरण के लिए,  
  c ^ = (a & lt; & lt; i) के स्थान पर;   
 आप कर सकते हैं:  
  c [i] ^ = (a [i] & lt; & lt; i);   
 कृपया ध्यान दें कि क्योंकि ^ = पिछले मान पर निर्भर करता है, सी [i] आरंभ करने से पहले कुछ अर्थपूर्ण होनी चाहिए।

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