Possible memory leak?

[Giovanni Bajo]

Fredrik Lundh wrote:

> since you know the length, you can preallocate the list
>
>     def iters3(n):
>         L = [None]*n
>         for i in xrange(n):
>             L[i] = chr(i%64)
>         return "".join(L)
>
> or use a preallocated array
>
>     def iters4(n):
>         L = array.array("B", [0])*n
>         for i in xrange(n):
>             L[i] = i%64
>         return L.tostring()

Of course. I was just trying to make a point about string accumulation being
O(n) and not O(n^2). You can get better factors with other solutions like the
ones you propose, but even the "simple" implementation (and most readable and
Pythonic, IMHO) behaves "well" under CPython 2.4.
-- 
Giovanni Bajo