Count the number of leading zeros as follows:
int lz = 8;
while (i)
{
lz--;
i >>>= 1;
}
Of course, this supposes the number doesn't exceed 255, otherwise, you would get negative results.
Question
I need int 32
in binary as 00100000
or int 127
in binary 0111 1111
.
The variant Integer.toBinaryString
returns results only from 1.
If I build the for loop this way:
for (int i= 32; i <= 127; i + +) {
System.out.println (i);
System.out.println (Integer.toBinaryString (i));
}
And from binary numbers I need the number of leading zeros (count leading zeros (clz) or number of leading zeros (nlz)) I really meant the exact number of 0, such ex: at 00100000 -> 2 and at 0111 1111 - > 1
Solution 2
Count the number of leading zeros as follows:
int lz = 8;
while (i)
{
lz--;
i >>>= 1;
}
Of course, this supposes the number doesn't exceed 255, otherwise, you would get negative results.
OTHER TIPS
How about
int lz = Integer.numberOfLeadingZeros(i & 0xFF) - 24;
int tz = Integer.numberOfLeadingZeros(i | 0x100); // max is 8.
Efficient solution is int ans = 8-(log2(x)+1)
you can calculate log2(x)= logy (x) / logy (2)
public class UtilsInt {
int leadingZerosInt(int i) {
return leadingZeros(i,Integer.SIZE);
}
/**
* use recursion to find occurence of first set bit
* rotate right by one bit & adjust complement
* check if rotate value is not zero if so stop counting/recursion
* @param i - integer to check
* @param maxBitsCount - size of type (in this case int)
* if we want to check only for:
* positive values we can set this to Integer.SIZE / 2
* (as int is signed in java - positive values are in L16 bits)
*/
private synchronized int leadingZeros(int i, int maxBitsCount) {
try {
logger.debug("checking if bit: "+ maxBitsCount
+ " is set | " + UtilsInt.intToString(i,8));
return (i >>>= 1) != 0 ? leadingZeros(i, --maxBitsCount) : maxBitsCount;
} finally {
if(i==0) logger.debug("bits in this integer from: " + --maxBitsCount
+ " up to last are not set (i'm counting from msb->lsb)");
}
}
}
test statement:
int leadingZeros = new UtilsInt.leadingZerosInt(255); // 8
test output:
checking if bit: 32 is set |00000000 00000000 00000000 11111111
checking if bit: 31 is set |00000000 00000000 00000000 01111111
checking if bit: 30 is set |00000000 00000000 00000000 00111111
checking if bit: 29 is set |00000000 00000000 00000000 00011111
checking if bit: 28 is set |00000000 00000000 00000000 00001111
checking if bit: 27 is set |00000000 00000000 00000000 00000111
checking if bit: 26 is set |00000000 00000000 00000000 00000011
checking if bit: 25 is set |00000000 00000000 00000000 00000001
bits in this integer from: 24 up to last are not set (i'm counting from msb->lsb)