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Can a Java 5-Op Log2 (Int 32) bit execute a bit in Java?

Just to clarify this is NOT a matter of homework, as I have seen that such accusations spilled out in contrast to other beat-hacking issues:

However, I have a bit in C:

#include <stdio.h>

const int __FLOAT_WORD_ORDER = 0;
const int __LITTLE_END = 0;

// Finds log-base 2 of 32-bit integer
int log2hack(int v)
{
  union { unsigned int u[2]; double d; } t; // temp
  t.u[0]=0;
  t.u[1]=0;
  t.d=0.0;

  t.u[__FLOAT_WORD_ORDER==__LITTLE_END] = 0x43300000;
  t.u[__FLOAT_WORD_ORDER!=__LITTLE_END] = v;
  t.d -= 4503599627370496.0;
  return (t.u[__FLOAT_WORD_ORDER==__LITTLE_END] >> 20) - 0x3FF;
}

int main ()
{
  int i = 25; //Log2n(25) = 4
  int j = 33; //Log2n(33) = 5

  printf("Log2n(25)=%i!\n",
         log2hack(25));
  printf("Log2n(33)=%i!\n",
         log2hack(33));

  return 0;
}

I want to convert this to Java. So far I have:

public int log2Hack(int n)
    {
        int r; // result of log_2(v) goes here
        int[] u = new int [2]; 
        double d = 0.0;
        if (BitonicSorterForArbitraryN.__FLOAT_WORD_ORDER==
                BitonicSorterForArbitraryN.LITTLE_ENDIAN) 
        {
            u[1] = 0x43300000;
            u[0] = n;
        }
        else    
        {
            u[0] = 0x43300000;
            u[1] = n;
        }
        d -= 4503599627370496.0;
        if (BitonicSorterForArbitraryN.__FLOAT_WORD_ORDER==
                BitonicSorterForArbitraryN.LITTLE_ENDIAN)
            r = (u[1] >> 20) - 0x3FF;
        else
            r = (u[0] >> 20) - 0x3FF;

        return r;
    }

(Note that this is inside the bit sort class ...)

Anyway, when I run it for the same values ​​of 33 and 25, I get 52 in each case.

I know Java integers are signed, so I'm sure it has something to do with why this happens. Does anyone have any ideas how I can get this 5-op, 32-bit integer log 2 to work in Java?

P.S. , : http://graphics.stanford.edu/~seander/bithacks.html#IntegerLogIEEE64Float

+5
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Java, 31 - Integer(v).numberOfLeadingZeros()? __builtin_clz, .

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0

int double . Java Double.longBitsToDouble, Double.doubleToLongBits. Java (, , ) big-endian, .

However, my attempt to adapt your code to Java did not help. Java integer subscription can be a problem.

0
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