I would like to convert a number to a BitArray, with the resulting BitArray only being as big as it needs to be.
For instance:
BitArray tooBig = new BitArray(new int[] { 9 });
results in a BitArray with a length of 32 bit, however for the value 9 only 4 bits are required. How can I create BitArrays which are only as long as they need to be? So in this example, 4 bits. Or for the number 260 I expected the BitArray to be 9 bits long
CodePudding user response:
You can figure out all the bits first and then create the array by checking if the least significant bit is 1 or 0 and then right shifting until the number is 0. Note this will not work for negative numbers where the 32nd bit would be 1 to indicate the sign.
public BitArray ToShortestBitArray(int x)
{
var bits = new List<bool>();
while(x > 0)
{
bits.Add((x & 1) == 1);
x >>= 1;
}
return new BitArray(bits.ToArray());
}
CodePudding user response:
Assuming you are working with exclusively unsigned integers, the number of bits you need is equal to the base 2 logarithm of the (number 1), rounded up.
CodePudding user response:
Counting the bits is probably the easiest solution.
In JavaScript for example...
// count bits needed to store a positive number
const bits = (x, b = 0) => x > 0 ? bits(x >> 1, b 1) : b;