How does this execute?
def f(x):
return x>0 and (x%2)+f(x/2) or 0
x is an array, for instance: [1, 1, 1, 3]
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5It doesn't execute. You can't apply the modulus operator to a list and Python will tell you that with a TypeError exception.msw– msw2010-05-19 11:54:39 +00:00Commented May 19, 2010 at 11:54
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4 Answers
This code is broken. For starters, x>0 is always true. But x%2 and x/2 yield type errors.
Comments
Did you mean this?
$ python
Python 2.5.5 (r255:77872, Apr 21 2010, 08:40:04)
[GCC 4.4.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> def f(x):
... return x>0 and (x%2)+f(x/2) or 0
...
>>> f([1, 1, 1, 3])
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 2, in f
TypeError: unsupported operand type(s) for %: 'list' and 'int'
Comments
evaluation in return statement is no different from evaluation in any other place. if x is a list this whole thing makes no sense and raises TypeError. x should be a numeric for this to work.
If x is a number it would work as follows:
- evaluate
x>0statement - if it was
Truereturn(x%2)+f(x/2)part. Which, of course, recurses infinitely - if it was
Falsereturn0
Comments
The function recursively counts the number of 1's in the binary form of the number x.
Each time the function adds sums the lowest bit (either 1 or 0) with the bit count of a number without the last bit (dividing by 2 is like shifting right by 1), or with 0 if there are no more bits.
For example: The function will return 2 for 5 as an input (5 is 101 in binary) The function will return 3 for 13 as an input (13 is 1101 in binary) ...
