Unsure about an 'and' operation

Question

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.

score 10 · Accepted Answer · answered Oct 11 '21 at 15:40

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.

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