One Java-eque solution is to pass in an object that has the arguments and
build the validity logic into that object.
So you have
public class LimitsAndLevels{
float[] whatever = null;
float anotherOne = 0.0; // or maybe some other overloaded value
public float[] getWhatever(){
isValid
Lemme see now...
>i laid out grandiose plans for a new beginning only to have my words
>fall on deaf ears. Have we become so self absorbed as to care only for
>our status and ego and not for the community at whole?
So you proposed a grandiose plane that is a heck of a lot of work for the
peo
There are a few long strings, but have fun yourself with the pi digit searcher:
http://www.angio.net/pi/bigpi.cgi
Longest string I heard of was nine 6's in a row, so search for 6 and
see what you get.
- Original Message -
From: "Ian Kelly"
To: "Jeffrey Gayn
(pulls out doctorate in Math.) Take a circle and measure its diameter, then
circumference (coffee cans and string are helpful). Then
pi = Circumference/diameter
approximating that is hard. It turns out that even though it *looks* like a
nice fraction, the value that results is not (fractions of
Re: Question on multiple python environments in Apache
Why wouldn't you use multiple apache instances?
On Fri, Oct 22, 2010 at 11:28 PM, Jeffrey Gaynor < jgay...@ncsa.uiuc.edu >
wrote:
I have several different versions of a web app that run under Apache. The issue
is t
I have several different versions of a web app that run under Apache. The issue
is that I need to have several different configurations available under several
environments (such as Django, mod_python and plain vanilla mod_wsgi). Is there
a simple way to get several of these to be completely ind
As I indicated, generating such triples is easy. What you found is the edge
case that
2*i*j = 200 => 100 = i*j
so (i,j) = (100,1) or (50,2) (25,4), (20,5) or (10,10). The maximal value are i
= 100, j = 1. The other sides are
i^2 - j^2 = 10,000 - 1 =
i^2 + j^2 = 10,000 + 1 = 10,001
...a
What you want is to realize that all integer Pythagorean triples can be
generated by a pair of integers, (i,j), j < i. The values are just (* =
multiply, ^ = exponentiation)
a = 2*i*j
b = i^2 - j^2
c = i^2 + j^2 (hypotenuse)
So yes indeed a^2 + b^2 = c^2. This is a very ancient result, btw and
Yip. I'm using it and for the most part like it. But...
I used their Java IDE for years (it totally rocks, highly recommended), so I it
is very comfortable to use PyCharm.
One thing that bugs me in refactoring though is that renaming a method or
variable does not necessarily work. It's suppose
rote:
> On Wed, Jul 28, 2010 at 4:41 PM, Jeffrey
> Gaynor wrote:
>> Hi,
>>
>> I am making a first large project in python and am having quite a
>> bit of difficulty unscrambling various python versions and what
>> they can/cannot do. To wit, I must communicate with
Hi,
I am making a first large project in python and am having quite a bit of
difficulty unscrambling various python versions and what they can/cannot do. To
wit, I must communicate with certain services via https and am required to
perform certificate verification on them.
The problem is that
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