[Tutor] I don't understand why this program is said to be "not defined" when I test it.

[Bob Gailer]

Also:
Briefly describe problem 22
Show us what you did to test the program
It would be better to test for integer before doing a numeric comparison
There is no need ever to compare a boolean expression to true or false. All
you need is either if condition or if not condition.
A good short cut for a range test is
if 0<= x <= 6:
It is not necessary to surround a condition with parentheses.
Your code will be simpler if you initially set Bueller to true, then change
it to false as needed. Please excuse my spelling as I am dictating this.
Who would have thought so many comments could come from such a simple piece
of code.
On Oct 16, 2015 7:17 PM, "Alan Gauld" <alan.gauld at btinternet.com> wrote:

> On 16/10/15 23:31, zak nelson wrote:
>
>> I don't understand why this program is said to be "not defined" when I
>> test
>> it.
>>
>
> In future please be more precise in describing the problem,
> and that means including the full error message in the post.
> It contains a lot of useful information.
>
> I'm guessing it doesn't say that the program is not defined
> but it says the name buler is not defined? Is that correct?
>
> Notice the spelling?
> Now look at your code
>
> def problem22(aList):
>>      length=len(aList)
>>      if (length>6):
>>          bueler=False
>>      else:
>>          bueler=True
>>          for i in aList:
>>              if(i < 0 and i > 6)==False:
>>                  bueler=False
>>              else:
>>                  buler=True
>>                  if type(i)!=("integer"):
>>                      bueler=False
>>                  else:
>>                      bueler=True
>>      return bueler
>>
>
> But there are a couple of other issues in there too.
> Specifically this line:
>
>           if(i < 0 and i > 6)==False:
>
> The bit in parentheses can never be true because i can
> never be both <0 and >6 at the same time. So the if
> statement is always true and the next line is always
> executed.
>
> Also the line
>
>    if type(i)!=("integer"):
>
> Will never be true because the string 'integer' is not a type.
> You should use the type name:
>
>    if type(i) != int
>
> or compare to the type of a known value:
>
>    if type(i) != type(5):
>
> In the case of an int the first one would be better.
> A final refinement is that for these kinds of comparisons
> it probably reads better to use is/is not rather
> than the ==/!= symbols, so:
>
>    if type(i) is not int:
>
> HTH
>
> --
> Alan G
> Author of the Learn to Program web site
> http://www.alan-g.me.uk/
> http://www.amazon.com/author/alan_gauld
> Follow my photo-blog on Flickr at:
> http://www.flickr.com/photos/alangauldphotos
>
>
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