Non-idempotent Python

Question

Write a few lines of Python code, X, that does not reference any global variables, such that

def method():
    X
    print(a)

method()

prints 1 but

def method():
    X
    X
    print(a)

method()

prints 2.

---

So, I hate to be a stickler, but it seems like vars and locals are actually global variables in Python:

def test_global_1():
    global vars, locals
    vars = lambda: 2
    locals = lambda: 3

def test_global_2():
    print(vars())
    print(locals())

test_global_1()
test_global_2()

---

Also, it looks like people would like to see objective winning criteria for puzzles like this. Code length doesn't really feel right here so perhaps we could make a system of brownie points for various novel features of the code? I'm not sure exactly what these could be but here's a start:

• +1 for really truly no globals (no vars or locals)
• +1 for being the first to post a particular technique
• +1 for the shortest solution posted
• +1 for a solution involving only a single Python statement
• +1 for interesting "hacks" like joining at lexical non-boundaries
• +1 for not using exceptions

And if you can think of more you could edit this question to add to the list.

Can this problem be solved without using exceptions and without using globals like vars and locals? I suspect it can, though I haven't figured out exactly how yet...

Answer 1

def method():
    if 'a' not in vars():a=0
    a+=1
    if 'a' not in vars():a=0
    a+=1
    print(a)

Initializes the variable a to 0 only if it's not already initialized in the variables table. Then, increments it.

More briefly (thanks to histocrat for len):

def method():
    a=len(vars())+1
    a=len(vars())+1
    print(a)

If the two copies of X could be on the same line, we could do

a=0;a+=1;a

which doubles to

a=0;a+=1;aa=0;a+=1;a

with the "sacrificial lamb" aa eating up the second variable assignment.

Answer 2

Python

Thought of this solution, since try and except was the first way I thought of to determine if a variable existed yet or not.

def method():
    try:a+=1
    except:a=1
    print(a)

Answer 3

Python 2

def method():
    exec'';locals()['a']=locals().get('a',0)+1
    exec'';locals()['a']=locals().get('a',0)+1
    print a

method()

Basically, when exec is encountered in Python 2, it causes a special flag (0x01) to be removed from method.func_code.co_flags, which makes locals assignments have an effect. I exploited this to implement nonlocal support in Python 2 (see line 43 for the xor that modifies the flag).

Answer 4

My first idea (and then smooshing it) was:

def method():
    a=2if'a'in vars()else 1 
    a=2if'a'in vars()else 1 
    print(a)

But histocrat's answer seems optimal.

Answer 5

def method(a=[]):  
  a.append(a)  
  print len(a)

Edited in response to comment: a is a list of empty lists of length n, where n is the number of time you've called method. Calling this method twice prints 1 then 2.

Answer 6

My attempt. Uses the math module to track if X is run once or twice.

def module():
  import sys
  if 'math' in sys.modules:
    a+=1
  else:
    a=1
  import math

  import sys
  if 'math' in sys.modules:
    a+=1
  else:
    a=1
  import math

  print(a)

module()

Answer 7

def method():
    #### X-block
    try:a
    except NameError:a=1
    else:a=2
    ####
    print(a)

The try block checks if the variable a is defined.
If variable not defined,(this is only when X-block is present once) then NameError Exception is raised.
If variable is defined, (this is when X-block is present twice) then else will be entered .