Menu
This is probably pretty basic, but to save me an hour or so of grief can anyone tell me how you can work out the number of bits required to represent a given positive integer in Java?
e.g. I get a decimal 11, (1011). I need to get the answer, 4.
I figured if I could work out how to set all the bits other than the most significant bit to 0, and then >>> it, I'd get my answer. But... I can't.
688
Number of Bits in a Specific Decimal IntegerA positive integer n has b bits when 2b-1 ≤ n ≤ 2b – 1. For example: 29 has 5 bits because 16 ≤ 29 ≤ 31, or 24 ≤ 29 ≤ 25 – 1. 123 has 7 bits because 64 ≤ 123 ≤ 127, or 26 ≤ 123 ≤ 27 – 1.
Answer: In general, N bits (binary digits) are required to represent unique integer values. Eight bits are required to represent 256 unique values. Those values range in binary from 00000000 through 11111111, or 0 through 255 in decimal.
Whole numbers (integers) are usually represented with 4 bytes, or 32 bits. In the past, symbols (e.g., letters, digits) were represented with one byte (8 bits), with each symbol being mapped to a number between 0-255.
Well, the answer is pretty simple. If you have an int value:
int log2(int value) { return Integer.SIZE-Integer.numberOfLeadingZeros(value); } The same exists for Long...
[Edit] If shaving milliseconds is an issue here, Integer.numberOfLeadingZeros(int) is reasonably efficient, but still does 15 operations... Expanding a reasonable amount of memory (300 bytes, static) you could shave that to between 1 and 8 operations, depending on the range of your integers.
134
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
