java - set bits 1-3 in an int given an int with only bits 1-3 set as desired -

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  अहस्ताक्षरित int x = 0xdeadbeef; अहस्ताक्षरित int y = 0x00000006; अहस्ताक्षरित int z = 0xdeadbee7;  x  और  y  में  मानों से  z  मान की गणना कैसे करें    
 बिट्स 1-3 का  y  है  011  और मुझे बिट्स से प्रतिस्थापित किया जाना चाहिए  x  में मान के 1-3 बिट्स चाहिए 1-3 से  y  अन्य बिट्स को छोड़कर के रूप में वे हैं।   
 
 
 यह जैसे आप चाहिए:  
     
  •  मास्क ऑफ बिट्स 1-3 ताकि वे  नहीं सेट  
  •  या बिट्स के परिणाम 1- 3 y   
     ऐसा:  
      // 0xe == 8 + 4 + 2 (यानी बिट्स 1-3) z = (x & amp; 0xe) | (वाई एंड 0xe);   
     ध्यान दें कि  ~  बिटवर्ड  उलटा  ऑपरेटर है, इसलिए 3 बिट्स को साफ़ करने के लिए आप उन बिट सेटों के उलटा लेते हैं, और उपयोग करते हैं वर्तमान मान के साथ उस व्युत्क्रम के बिटवर्ड और।  
     आप केवल  | Y  के बजाय  | (Y & amp; 0xe)  यदि आप जानते हैं कि केवल बीट्स 1-3 को  y  में सेट किया जाएगा। (ध्यान दें कि मैंने मान लिया है कि "बिट्स 1-3 "का अर्थ है बिट्स 0, 0 से शुरू करने के सामान्य बिट-गिनती तंत्र का उपयोग करते हुए, मान 8, 4 और 2 के साथ बिट्स। यदि आप" नीचे तीन बिट्स "का मतलब है, जो आमतौर पर बिट्स 0-2 हो, डी कोड में 0x के बजाय 7 का उपयोग करें।)

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