On 9/19/07, Boykie Mackay <[EMAIL PROTECTED]> wrote: > Hi guys, > > I have come across a bit of code to find if a group of numbers is odd or > even.The code snippet is shown below: > > if not n&1: > return false > > The above should return false for all even numbers,numbers being > represented by n.I have tried to wrap my head around the 'not n&1' but > I'm failing to understand what's going on.Could someone please explain > the statement.
I'll respond more from a general programming point of view than as a Python expert (far from it!). You can always tell if a regular number is divisible by 10, by just looking at the last digit; if it's a zero, it is, and if it's anything else, it isn't. That's a property of the decimal (base 10) system. The above snippet is taking advantage of a similar property of binary numbers, which are base 2. What the above snippet is doing is checking to see if that last digit is a 0 or not (asking "not n&1" is equivalent to asking "n&0", since that digit can only be a 0 or 1). Since binary numbers are base 2, any binary number that ends in a 0 is divisible by 2, analogous to the decimal number example above. The technique being used to determine the value of the last digit is a bitwise-AND. It compares the binary digits in the representation of n with the binary digits in the representation of 1 on a one-to-one positional arrangement. If either number has a zero in a certain position, the result will have a zero in that position as well; if both numbers have a 1 in that position, the result will have a 1. As an example, I'll represent 4 as a binary number: 100 (4 in binary) 001 (1 in binary) --------- 000 (Result of bitwise-AND) and now using 5 as our input: 101 (5 in binary) 001 (1 in binary) -------- 001 (Result of bitwise-AND) Since 1 represented in binary is all zeroes except for the last digit, the only way to get anything different from 000 as a result is for the last digit to be 1. Hopefully this didn't confuse you anymore. Experts, feel free to chime in. Tony R.
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