Make a for loop repeat forever

Question

for loops are useful in python in many ways, but one of the most common usages associated with a for loop is it's ability to repeat a certain amount of times. Like this:

for times in range(50):
    print "I'm repeating this 50 times..."

The challenge I propose to you is to make a for repeat forever. If you can find a way to do it without using a recursion, kudos to you. You can use any major version of Python. Just specify which one you are using. The goal is to do this in the smallest file size you can get, and a method that won't "break" after a while. Good luck!

Answer 1

Because this was turned to a codegolf, have a 48 bytes version as a one-liner. 3 bytes could be removed from both by replacing print i with pass.

for i in __import__('itertools').count():print i

Or a 47 bytes version with two lines.

from itertools import*
for i in count():print i

---

Old message:

Easy it is.

def f(i = 0):
    while True:
        yield i
        i += 1
for i in f():
    print(i)

Answer 2

This can be done quite easily by abusing floating point addition.

def frange(start, stop, step):
    r = start
    while r < stop:
        yield r
        r += step

for x in frange(0, 1e16, 1):
    print (x)

IEEE754 64-bit floats have 53 bits of mantissa (including the implicit bit), so if x > 1<<53 then x + 1 == x. 1 << 53 is approximately 9e15.

Answer 3

This is an interesting way of doing it (32 chars for class x, 46 not counting the print):

class x:__iter__=next=lambda s:s
for i in x():print "Hello"

This defines an iterable whose next method never raises a StopIteration

Answer 4

I don't usually golf in Python, probably some reason people haven't already done posted this 23-character solution.

x=[1]
for i in x:
 x+=x

Much less memory-hungry if you do x+=[i] instead. Either one hangs my 2.7.5

Answer 5

for i in iter(lambda : 0, 1): 
    print (i) # prints 0 forever 

The iter function is normally used to create an iterator from an iterable.

The iter function can also take two arguments, a callable and a sentinel value. In that case it returns an iterator that yields the result of the callable until the callable returns the sentinel value.

In this case, the callable is a lambda that always returns 0. Since it never can return the sentinel value (1), this will loop forever.

Since this is code golf, this version is 28 characters and only one line:

for i in iter(lambda:0,1):i

Answer 6

28 bytes, infinite, and no itertools

for x in iter(int,1):print 1

Shouldn't end without system program cutoffs (i.e. not my fault).

22 bytes without printing anything:

for x in iter(int,1):1

Answer 7

Itertools, to the rescue:

from itertools import count
for i in count():
    print i

or

from itertools import repeat
for i in repeat(42):
    print i

Answer 8

39 Characters, no itertools and no crashing

def f():
 while 1:yield
for i in f():1

Itertools is unnecessary, you can define your own generator.

Answer 9

Python 2 & 3: 37 characters

class P:__getitem__=id
for t in P():1

¡No infinite memory required!

Answer 10

Python 3.3: 2 lines, 29 characters

j=[1]  
for i in j: j.append(1)

Does this break? I've run it for a short while, doesn't seem to hang.