Alan,
Wow! Those changes radically change the program!
Is there a way to get the try-exception to cause the end user questions
to be repeated until acceptable answers are given? Like a kind of
"return"
or "repeat" command that could be added to the except side of things?
try:
start = int(raw_input("Enter yyyy for beginning year : "))
end = int(raw_input("Enter yyyy for ending year : "))
except ValueError: print "Must be a integer number as YEAR."
I also noticed that I had to make the statement "start = end = 0" prior
to the 'try'. To my eye, it looks to me that 'start' and 'end' are
being
declared inside the try-exception, but evidently python does not agree.
Thanks,
Terry
On Tue, 2007-07-03 at 21:42 +0100, Alan Gauld wrote:
> "Terry" <[EMAIL PROTECTED]> wrote
>
> > trapping for possible user errors.
> > I am using x.isalnum() to check that I have a number
>
> First, alnum() checks for alpha-numeric characters so
> will allow both alphabetic characters and numerics.
>
> BUT in python its better to ask forgiveness that permission
> so don't bother doing the check but catch the errors when
> they occur:
>
> try: start = int(start); end = int(end)
> except: ValueError: start = None # or whatever you prefer
>
> > def leap(yyyy):
> > answer = 0
> > t1 = yyyy / 4
> > if t1 == int(t1):
>
> For integer division this will always be true.
>
> > t2 = yyyy / 100
> > t3 = yyyy / 400
> > if t2 <> int(t2) or t3 == int(t3):
>
> and similarly the second test will always be true.
>
> > answer = "-- leap year!"
>
> thus this will always be true.
>
> > start = raw_input("Enter yyyy for beginning year : ")
> > end = raw_input("Enter yyyy for ending year : ")
>
> The easiest way to solve your conversion problems
> is to do it at source so wrap raw_input in a call to int()
> and wrap both prompts in a try/except as shown above.
>
> try:
> start = int(raw_input....)
> end = int(raw_input....)
> except ValueError: pass # do something with error
>
> > if len(start) == 4 and len(end) == 4:
>
> that way you don't need these lines but can instead
> test if the numeric value is greater than some lower
> limit
>
> > if start.isalnum() == 1 and end.isalnum() == 1:
> > #<----------fails to detect 'aaaa' as not a number
> > start = int(start); end = int(end)
> > #<----------commits suicide here
> > if start > 0 and end > start:
>
> As you are doing it start can never be less than 1000
> (unless the user enters 0001 I suppose...)
>
> Also in Python you can use the neater style of
>
> if 0 < start < end:
>
> to test whether start lies between 0 and end.
>
> > for i in range(start, end + 1):
> > answer = leap(i)
> > if answer != 0:
>
> Its best not to mix return types in a function. If it is a
> leap year you return a string. If it's not, you return a
> number - 0. It would probably be better for the function
> to return a boolean: True if it is a leap year and False
> if not. That style of function is often called a predicate
> and is usually named with an 'is' in front so it would
> become isLeapYear() Then this test becomes the more
> readable
>
> for y in range(start, end+1): # use y for year instead of meaningless
> i
> if isLeapYear(y):
> print i, 'is a leap year'
>
> HTH,
>
>
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