Please don't top-post. Put your response under the quote you're
responding to. And trim the parts that are no longer relevant. I've
rearranged this message to try to pretend that you did that.
> On Wed, Nov 26, 2014 at 3:46 PM, Alan Gauld <alan.ga...@btinternet.com>
> wrote:
>
>> On 26/11/14 09:57, Sunil Tech wrote:
>>
>>> Hi Danny,
>>>
>>> Curious to the use the need of using while True in the given example of
>>> ask_for_a_digit().
>>>
>>>
>>> On Mon, Nov 17, 2014 at 9:57 AM, Danny Yoo <d...@hashcollision.org
>>> <mailto:d...@hashcollision.org>> wrote:
>>>
>>> > def ask_for_a_digit():
>>> > while True:
>>> > digit = raw_input("Give me a digit between 0 and 9.")
>>> > if digit not in "0123456789":
>>> > print "You didn't give me a digit. Try again."
>>> > else:
>>> > return int(digit)
>>>
>>
>> The while loop makes it keep on asking until a valid input is
>> received. Without the while loop it would only ask once and
>> either return None or a digit.
>>
On 11/26/2014 06:16 AM, Sunil Tech wrote:
Thank you Alan. But a question here, how would it understand that the given
input is valid?
Inside the while loop there is a else clause containing a return
statement. That's how the code escapes the while loop: whenever the
user enters something deemed correct.
More traditionally, a break will exit a loop. Or the while can contain
a more complex condition. Example of that, untested:
def ask_for_a_digit()
digit = "xxx"
while len(digit) != 1 or digit not in "0123456789":
digit = raw_input("Give me a digit between 0 and 9.")
Unfortunately, this form doesn't include the "chiding" of the user.
That's more painful, but it can be done.
def ask_for_a_digit()
digit = ()
while len(digit) != 1 or digit not in "0123456789":
if digit = (): print "You didn't give me a digit. Try again"
digit = raw_input("Give me a digit between 0 and 9.")
--
DaveA
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