binary - Obtaining reversed Integer in Java -


There is a method in Java that reverses the bit in an integer reversebits (). I wanted to try another implementation and I have it: 

  public static in reverse (int no) {int num_rev = 0; For (Int i = 0; I > I) & amp; 1); If (((>>) & amp; 1)! = 0) {num_rev = num_rev | (Intel) Math.Po (2, Integer. SSE-I); }} Return num_rev; }   
 The result is not num_rev correct. Anyone have any idea how to "rebuild" the value? Perhaps there is a better way to do this?  
 Thanks for any suggestions   
 
 
 Common ways to reverse the bit Will be through bit manipulation, and definitely  no  floating point math routine!  
 For example (nb: untested).  
  int reverse (int x) {int y = 0; For (int i = 0; i <32; ++ i) {y & lt; & Lt; = 1; // mer space y = (X & 1); // copy LSB's x in the x> & gt; & Gt; = 1; // shift X correct} return y; }   
 because  x  is moved right and  y  moved to the left that the  x  Eventually MSB becomes  y .  
 A good (and well-known method) method is:  
  unsigned infinite reverse (unsigned int x) {x = ((((x and xxaaaaaaa ) & Gt; & gt; 1) | ((x and 0x55555555)  gt; & gt; 16) | (x & lt; & lt; 16);) }   
 This is actually a C code, but as in Java, the  unsigned  type of port is not in Java, by removing what you should do < Code> unsigned  abilities and  & gt; & Gt; & Gt;  & gt; instead of  & Gt; Use  to make sure that you do not get any "sign extensions".  
 It changes every other  bit , then every other  pair  of bits, then every other  nybble , then every Second  byte , and then finally the top and bottom 16-bit word It really works :)

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