Forum: Ruby Array troubles

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(Guest)
on 2007-05-25 18:31
(Received via mailing list)
I'm converting a Java calculator program I wrote into Ruby--which,
just so you know, has been unbelievable with how quickly the language
can be learned and implemented and how easy it made programming what
had taken months of Java (but that was during school-time so
development was, admittedly, very slow)--but I've run into a problem.

In a term, if you have q/q^2/4/8, it turns into q^-1/.5, but I'm
getting q^-10! I've narrowed it down to one part of one function:
division in Term#simplify.

There's a lot of code involved, so I'll only post relevant data. I
REALLY do think you'll need any more than this. Also: when I say
algebraic calculator, I mean that half the time you're working with
letters (variables), not numbers.

Term#simplify

>>                num1 = @items[count-1]
>>                        temp << num1/num2
>>                        puts 'temp post ' + (temp).to_s
>>                    end
>>                end
>>            end
>>            count += 1
>>        end
>>        puts temp[1]
>>        temp
>>    end

With q/q^2/4/8 it prints out

>> temp pre
>> q/q^2 = q^-1
>> temp post q^-1
>> temp pre q^-1
>> 4/8 = 0
>> temp post q^-10
>> 0

Interestingly enough, 4/8=0. I don't know how the hell that works.
Also, when it goes around for the second time to get 4/8, it puts it
in the first place in the array (array[0]) and not the next space,
which is where I thought this method is supposed to put it. ri must
have this wrong too.

I don't think you need to know how I divide variables, but just in
case you do:

>>    # Return the result of this instance and var1.
>>    # Assumes that the two are like.
>>    def /(var)
>>        Variable.new(self.base, (self.exponent-var.exponent))
>>    end

A variable is just a string base and integer exponent.

PS: I would accept some criticism as long as it's productive, but I do
know that my code is lacking in certain areas (like how it doesn't
check whether or not the variables have like bases), but it is a WIP.
Robert K. (Guest)
on 2007-05-25 18:51
(Received via mailing list)
On 25.05.2007 16:26, removed_email_address@domain.invalid wrote:
> There's a lot of code involved, so I'll only post relevant data. I
>>>
>>>                    # You can't do much if the operands aren't of the same class.
>>>                    if num1.class.eql?( num2.class )
>>>                        puts 'temp pre ' + (temp).to_s
>>>                        # And use the / method of whatever class they are.
>>>                        puts num1.to_s + item + num2.to_s + " = " +
>>>                            (num1/num2).to_s

You don't need all the #to_s, instead you can do

print num1, item, num2, " = ", (num1/num2), "\n"

or use printf or puts with string interpolation.

>
> Interestingly enough, 4/8=0. I don't know how the hell that works.
That's standard behavior for integer math in programming languages.  You
might want to look into mathn / Rational:

irb(main):001:0> 4/8
=> 0
irb(main):002:0> require 'mathn'
=> true
irb(main):003:0> 4/8
=> 1/2
irb(main):004:0> (4/8).class
=> Rational
irb(main):005:0> (4/8).to_f
=> 0.5
irb(main):006:0> "%10f" % (4/8)
=> "  0.500000"

>>>    def /(var)
>>>        Variable.new(self.base, (self.exponent-var.exponent))
>>>    end
>
> A variable is just a string base and integer exponent.
>
> PS: I would accept some criticism as long as it's productive, but I do
> know that my code is lacking in certain areas (like how it doesn't
> check whether or not the variables have like bases), but it is a WIP.

Hope the math hint gets you started.

Kind regards

  robert
(Guest)
on 2007-05-25 19:18
(Received via mailing list)
> Hope the math hint gets you started.
>
> Kind regards
>
>         robert

Thanks bundles of bundles! You haven't solved my main problem (the
term array), but this is damn illuminating. Also, it might help me
when I'm trying to make this calc with fractions because this already
does fractions! The problem will just be making them compatible with
variables.
Dan Z. (Guest)
on 2007-05-25 20:22
(Received via mailing list)
removed_email_address@domain.invalid wrote:
> In a term, if you have q/q^2/4/8, it turns into q^-1/.5, but I'm
> getting q^-10!

Hmm... shouldn't that be q^-1/32 ?
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