Ok, I'm relatively new to Python (coming from C, C++ and Java). I'm
working on a program that outputs text that may be arbitrarily long,
but should still line up, so I want to split the output on a specific
column boundary. Since I might want to change the length of a column,
I tried defining the column as a constant (what I would have made a
"#define" in C, or a "static final" in Java). I defined this at the
top level (not within a def), and I reference it inside a function.
Like this:
COLUMNS = 80
def doSomethindAndOutputIt( ):
...
for i in range( 0, ( len( output[0] ) / COLUMNS ) ):
print output[0][ i * COLUMNS : i * COLUMNS + ( COLUMNS - 1 ) ]
print output[1][ i * COLUMNS : i * COLUMNS + ( COLUMNS - 1 ) ]
..
etc. etc. It works fine, and splits the output on the 80-column
boundary just like I want.
Well, I decided that I wanted "COLUMNS = 0" to be a special "don't
split anywhere" value, so I changed it to look like this:
COLUMNS = 80
def doSomethindAndOutputIt( ):
...
if COLUMNS == 0:
COLUMNS = len( output[ 0 ] )
for i in range( 0, ( len( output[0] ) / COLUMNS ) ):
print output[0][ i * COLUMNS : i * COLUMNS + ( COLUMNS - 1 ) ]
print output[1][ i * COLUMNS : i * COLUMNS + ( COLUMNS - 1 ) ]
..
Now, when I run it, I get the following error:
Traceback (most recent call last):
File "Test.py", line 140, in ?
doSomethingAndOutput( input )
File "Test.py", line 123, in doSomethingAndOutput
if COLUMNS == 0:
UnboundLocalError: local variable 'COLUMNS' referenced before
assignment
I went back and re-read chapter 13 of "Learning Python", which talks
about variable scoping rules, and I can't figure out why Python is
saying this variable in Unbound. It works if I insert:
global COLUMNS
before the "if" statement... but I don't understand why. Is the
interpreter scanning my entire function definition before executing
it, recognizing that I *might* assign COLUMNS to a value, and deciding
that it's a local on that basis?
--
http://mail.python.org/mailman/listinfo/python-list
