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comp.lang.ruby

Array troubles

Hakusa@gmail.com

5/25/2007 2:26:00 PM

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

>>def simplify
>> count=0 # count will keep track of iterations.
>> temp = [] # This array will be our temp for @items.
>> @items.each do |item|
>>
>> # Why check for this?
>> # The only time strings are use here is as an operator.
>> if item.class == String
>> # Defining these really just make the code easier for further on.
>> # They're the items before and after the operator.
>> num1 = @items[count-1]
>> num2 = @items[count+1]
>>
>> # So if the operator is division...
>> if item.eql?('/')
>> # 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
>> 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.

3 Answers

Robert Klemme

5/25/2007 2:48:00 PM

0

On 25.05.2007 16:26, Hakusa@gmail.com wrote:
> 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
>
>>> def simplify
>>> count=0 # count will keep track of iterations.
>>> temp = [] # This array will be our temp for @items.
>>> @items.each do |item|
>>>
>>> # Why check for this?
>>> # The only time strings are use here is as an operator.
>>> if item.class == String
>>> # Defining these really just make the code easier for further on.
>>> # They're the items before and after the operator.
>>> num1 = @items[count-1]
>>> num2 = @items[count+1]
>>>
>>> # So if the operator is division...
>>> if item.eql?('/')
>>> # 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.

>>> 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.

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"

> 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.

Hope the math hint gets you started.

Kind regards

robert

Hakusa@gmail.com

5/25/2007 3:10:00 PM

0

> 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 Zwell

5/25/2007 4:21:00 PM

0

Hakusa@gmail.com 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 ?