Python cannot handle numbers string starting with 0. Why?

user796530 picture user796530 · Oct 22, 2012 · Viewed 9.3k times · Source

I just executed the following program on my python interpreter:

>>> def mylife(x):
...     if x>0:
...             print(x)
...     else:
...             print(-x)
... 
>>> mylife(01)
File "<stdin>", line 1
mylife(01)
        ^
SyntaxError: invalid token
>>> mylife(1)
1
>>> mylife(-1)
1
>>> mylife(0)
0

Now, I have seen this but as the link says, the 0 for octal does not work any more in python (i.e. does not work in python3). But does that not mean that the the behaviour for numbers starting with 0 should be interpreted properly? Either in base-2 or in normal base-10 representation? Since it is not so, why does python behave like that? Is it an implementation issue? Or is it a semantic issue?

Answer

mgilson picture mgilson · Oct 22, 2012

My guess is that since 012 is no longer an octal literal constant in python3.x, they disallowed the 012 syntax to avoid strange backward compatibility bugs. Consider your python2.x script which using octal literal constants:

a = 012 + 013

Then you port it to python 3 and it still works -- It just gives you a = 25 instead of a = 21 as you expected previously (decimal). Have fun tracking down that bug.