I came across a piece of code to calculate the number of binary bits needed for a decimal.
nbits = 1 + (decimal and floor(log2(decimal)))
I understand that 1+floor(log2(decimal)) returns the number of nbits.
However I'm not sure what the and statement ensures here.
It's taking advantage of the fact that 0 is a falsey value; it's a compact form of
bits = 1 + (0 if decimal == 0 else floor(log2(decimal)))
or even less compactly,
if decimal == 0:
bits = 1 # 1 + 0
else:
bits = 1 + floor(log2(decimal)))
floor(log2(0)) is undefined, so you need to handle decimal == 0 specially.
x and y == y for any truthy value of x, and x and y == x (without evaluating y at all) for a falsey value.
In short, it says that bits is at least 1 bit (0), but may require additional bits for non-zero values.