continue statement: best way round a problem

[Skip Montanaro]

    Steve> I'm trying to write a processing loop which, when any of a number
    Steve> of error conditions occurs, proceeds to the next iteration.  This
    Steve> looked like an obvious case for continue, so my code reads (with
    Steve> extraneous elements removed) as follows:
    ... [snip] ...
    Steve> As the advanced reader may have already predicted, I get a syntax
    Steve> error with the second continue:

    Steve> SyntaxError: 'continue' not properly in loop

I think that's a bug in the Python parser.  Clearly, the second continue
statement is in a valid place.  It's inside the for loop and not executed
from a try clause.  The syntax error seems to be related to the complexity
of the try/except statements.

Here's a workaround:

    for i in range(10):
	print i
	try:
	    try:
		if i > 6:
		    1/0
	    except ZeroDivisionError:
		raise "continue"
	    print "a"
	    try:
		if (i+1) % 3 == 0:
		    print "c"
		    1/0
	    except KeyboardInterrupt:
		raise
	    except:
		raise "continue"
	    print "b"
	except KeyboardInterrupt:
	    print "Keyboard Interrupt"
	    break
	except "continue":
	    continue

If you replace the 'raise "continue"' statements with continue statements
you'll get the syntax error.

-- 
Skip Montanaro (skip at mojam.com)
http://www.mojam.com/
http://www.musi-cal.com/