compute the double square...... :(

[aregee]

hey all thanks for yr help,i got it right .....n sorry for
confussions...i m very new to python...just started learning it couple
of days ago...

Ian Kelly wrote:
> On 1/8/2011 11:10 PM, aregee wrote:
> > pie.py:3: Deprecation Warning: integer argument expected, got float
> >    for b in range(0,(x**0.5)/2):
>
> I expect you want range(0, int((x / 2) ** 0.5) + 1), no?
>
> > for b in range(0,(x**0.5)/2):
> >        a = (x-(b**2))**0.5
> > try:
> >        a = int(a)
> > except:
> >        print("not an integer")
> >        exit(1)
>
> Your indentation is confusing.  Is the try-except contained inside the
> for loop or not?
>
> And what are you actually trying to test for here?  The assignment here
> of "a = int(a)" will never throw an exception as long as the loop runs.
>
> >
> >        count = 0;
> >        count = count + 1;
>
> Again, confusing indentation.  Is this supposed to be part of the except
> block?  And what is the purpose of incrementing count if you're going to
> set it to 0 immediately before?  You might as well just write "count = 1"
>
> > if (x == a**2 + b**2):
> >
> >        print "double square"
>
> This also appears to be outside of the loop.