LinuxQuestions.org - python overflow error

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python overflow error

I keep getting
OverflowError: Python int too large to convert to C long
on a python script when I run it on 32 bit systems but it works as expected on 64bit

Code:

for number in xrange(int("0101010101"), int("9898989898")+1):

        number = `number`.zfill(10)

        prev = ''

        i = 0

        for c in `number`:

                if c in prev:

                        break

                else:

                        i += 1

                        prev = c

                if i>=len(`number`):

                        print number

Here is a perl equivalent which works on both 32bit and 64bit

Code:

my $num = "";

for $num ("0101010101" .. "0101989898"){

        if ($num =~ /00|11|22|33|44|55|66|77|88|99/o) {

                ++$num;}

        else {

                print "$num\n";}}

I'm trying to use this on a raspberry pi which is 32bit I need to do this in python if posible. Is there an easy way to get around this error?

I would make something like this:

Code:

for number in xrange(int("0101010101"), int("9898989898")+1):

  number = `number`.zfill(10)

  n = str(number)

  found = 0

  for c in range(1, len(n) -1):

    found += n[c] == n[c+1]

  if not found:

    print number

Quote:

Originally Posted by pan64 (Post 4918173)

I would make something like this:

Code:

for number in xrange(int("0101010101"), int("9898989898")+1):

  number = `number`.zfill(10)

  n = str(number)

  found = 0

  for c in range(1, len(n) -1):

    found += n[c] == n[c+1]

  if not found:

    print number

Thanks for the suggestion but still get overflow errors with your code. The problem seems to be with xrange. I ended up doing this which is really messy but works.

Code:

import itertools



for number in itertools.count(int("0101010101"), 1):

        number = `number`.zfill(10).strip('L')

        prev = ''

        i = 0

        if number > "9898989898":

                break

        for c in `number`:

                if c in prev:

                        break

                else:

                        i += 1

                        prev = c

                if i>=len(`number`):

                        print number

The .strip('L') is there because for some reason python after counting past 3000000000 starts adding a L to the end.