Possible memory leak?

[Fredrik Lundh]

Giovanni Bajo wrote:

> ------- foo.py -----
> def iters(n):
>     s = ''
>     for i in xrange(n):
>         s += chr(i%64)
>     return s
>
> def iters2(n):
>     L = []
>     for i in xrange(n):
>         L.append(chr(i%64))
>     return "".join(L)
> ------- foo.py -----
>
> So, look, it's even faster than the solution you're proposing.

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()

on my machine, the last one is twice as fast as your "even faster"
solution under 2.4.  in earlier versions, it's just under 5 times faster.

for the OP's problem, a PIL-based solution would probably be ~100
times faster than the array solution, but that's another story.

</F>