On Sunday, October 9, 2016 at 2:41:41 PM UTC+1, BartC wrote:
> On 09/10/2016 13:01, Cai Gengyang wrote:
> > I'm moving on to chapter 9
> > (http://programarcadegames.com/index.php?lang=en&chapter=lab_functions) of
> > programarcadegames for the time being and going back to chapter 8 later
> > (its just fucking frustrating and doesn't seem to work for the time being
> > and I have a very bad temper).
> >
> > At least for chapter 9, I got the first part correct at first try ---
> > define a function that takes and prints the smallest of 3 numbers. I pasted
> > it here for reference and discussion. The code can also be further modified
> > to include a clause that says that if two numbers tie for smallest, choose
> > any of the two numbers. (I will attempt to do it and then post it here
> > again for future discussion with users here
> >
> >
> >>>> def min3(a, b, c):
> > min3 = a
> > if b < min3:
> > min3 = b
> > if c < min3:
> > min3 = c
> > if b < c:
> > min3 = b
> > return min3
> >
> >>>> print(min3(4, 7, 5))
> > 4
> >
>
> The exercise says you must use an if/elif chain. The first thing that
> comes to mind is:
>
> def min3(a,b,c):
> if a<=b and a<=c:
> return a
> elif b<=a and b<=c:
> return b
> else:
> return c
>
> The bit about numbers tying for smallest is not meaningful; the caller
> can't tell if a minimum value of 42 came from a, b or c.
>
> --
> Bartc
The Pythonic way
if b >= a <= c:
...
Kindest regards.
Mark Lawrence.
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