def calculate (options , *numbers)
options << "add" if options.empty?
numbers.inject(0) { |sum, number| sum + number } if options == "add"
numbers.inject() { |sum, number| sum - number } if options == "subtract"
end
But now I see this output in Rubymonks :
defaults to addtion when no option is specifiedNoMethodErrorundefined method `empty?' for 4:Fixnuminvoking calculate(4, 5, add: true) returns 9NoMethodErrorundefined method `empty?' for 4:Fixnuminvoking calculate(-10, 2, 3, add: true) returns -5NoMethodErrorundefined method `empty?' for -10:Fixnuminvoking calculate(0, 0, 0, 0, add: true) returns 0NoMethodErrorundefined method `empty?' for 0:Fixnum
I think the problem is that I cannot take care that the right values are in the right variable.
Anyone a tip how to solve this ?
Roelof
The options hash is the first parameter to your function, so pass it
as the first parameter:
calculate({:add => true}, 4,5)
and also, you shouldn't check options == "add", but options[:add] ==
true or just options[:add]
Jesus.
···
On Fri, Sep 28, 2012 at 9:40 AM, Roelof Wobben <rwobben@hotmail.com> wrote:
Hello,
I have this script:
def calculate (options , *numbers)
options << "add" if options.empty?
numbers.inject(0) { |sum, number| sum + number } if options == "add"
numbers.inject() { |sum, number| sum - number } if options == "subtract"
end
But now I see this output in Rubymonks :
defaults to addtion when no option is specifiedNoMethodErrorundefined method
`empty?' for 4:Fixnuminvoking calculate(4, 5, add: true) returns
9NoMethodErrorundefined method `empty?' for 4:Fixnuminvoking calculate(-10,
2, 3, add: true) returns -5NoMethodErrorundefined method `empty?' for
-10:Fixnuminvoking calculate(0, 0, 0, 0, add: true) returns
0NoMethodErrorundefined method `empty?' for 0:Fixnum
I think the problem is that I cannot take care that the right values are in
the right variable.
Date: Fri, 28 Sep 2012 17:09:08 +0900
From: jgabrielygalan@gmail.com
Subject: Re: calcaulation with unknown numbers of numbers and options fail
To: ruby-talk@ruby-lang.org
> Hello,
>
> I have this script:
>
> def calculate (options , *numbers)
> options << "add" if options.empty?
> numbers.inject(0) { |sum, number| sum + number } if options == "add"
> numbers.inject() { |sum, number| sum - number } if options == "subtract"
> end
>
> But now I see this output in Rubymonks :
>
> defaults to addtion when no option is specifiedNoMethodErrorundefined method
> `empty?' for 4:Fixnuminvoking calculate(4, 5, add: true) returns
> 9NoMethodErrorundefined method `empty?' for 4:Fixnuminvoking calculate(-10,
> 2, 3, add: true) returns -5NoMethodErrorundefined method `empty?' for
> -10:Fixnuminvoking calculate(0, 0, 0, 0, add: true) returns
> 0NoMethodErrorundefined method `empty?' for 0:Fixnum
>
> I think the problem is that I cannot take care that the right values are in
> the right variable.
>
>
> Anyone a tip how to solve this ?
The options hash is the first parameter to your function, so pass it
as the first parameter:
calculate({:add => true}, 4,5)
and also, you shouldn't check options == "add", but options[:add] ==
true or just options[:add]
Jesus.
Thanks,
I can't change the function call because I try to learn ruby by a interactive website.
I don't understand one remark. You say I shouldn't check the options =="add" .
What I' trying to do is to add the word "add" to options if the options array = empty.
So the function defaults to "add" when nothing is entered.
Roelof
···
On Fri, Sep 28, 2012 at 9:40 AM, Roelof Wobben <rwobben@hotmail.com> wrote:
Date: Fri, 28 Sep 2012 17:09:08 +0900
From: jgabrielygalan@gmail.com
Subject: Re: calcaulation with unknown numbers of numbers and options fail
To: ruby-talk@ruby-lang.org
> Hello,
>
> I have this script:
>
> def calculate (options , *numbers)
> options << "add" if options.empty?
> numbers.inject(0) { |sum, number| sum + number } if options == "add"
> numbers.inject() { |sum, number| sum - number } if options == "subtract"
> end
>
> But now I see this output in Rubymonks :
>
> defaults to addtion when no option is specifiedNoMethodErrorundefined
> method
> `empty?' for 4:Fixnuminvoking calculate(4, 5, add: true) returns
> 9NoMethodErrorundefined method `empty?' for 4:Fixnuminvoking
> calculate(-10,
> 2, 3, add: true) returns -5NoMethodErrorundefined method `empty?' for
> -10:Fixnuminvoking calculate(0, 0, 0, 0, add: true) returns
> 0NoMethodErrorundefined method `empty?' for 0:Fixnum
>
> I think the problem is that I cannot take care that the right values are
> in
> the right variable.
>
>
> Anyone a tip how to solve this ?
The options hash is the first parameter to your function, so pass it
as the first parameter:
calculate({:add => true}, 4,5)
and also, you shouldn't check options == "add", but options[:add] ==
true or just options[:add]
Jesus.
Thanks,
I can't change the function call because I try to learn ruby by a
interactive website.
Then you have to modify how you declare the function.
I don't understand one remark. You say I shouldn't check the options
=="add" .
What I' trying to do is to add the word "add" to options if the options
array = empty.
So the function defaults to "add" when nothing is entered.
Options is not an array, it's a hash. but in any case it's never going
to be equal to "add". It will contain an element or a key "add".
Jesus.
···
On Fri, Sep 28, 2012 at 10:18 AM, Roelof Wobben <rwobben@hotmail.com> wrote:
On Fri, Sep 28, 2012 at 9:40 AM, Roelof Wobben <rwobben@hotmail.com> >> wrote:
Date: Fri, 28 Sep 2012 17:31:55 +0900
From: jgabrielygalan@gmail.com
Subject: Re: calcaulation with unknown numbers of numbers and options fail
To: ruby-talk@ruby-lang.org
>
>
>> Date: Fri, 28 Sep 2012 17:09:08 +0900
>> From: jgabrielygalan@gmail.com
>> Subject: Re: calcaulation with unknown numbers of numbers and options fail
>> To: ruby-talk@ruby-lang.org
>
>>
>> > Hello,
>> >
>> > I have this script:
>> >
>> > def calculate (options , *numbers)
>> > options << "add" if options.empty?
>> > numbers.inject(0) { |sum, number| sum + number } if options == "add"
>> > numbers.inject() { |sum, number| sum - number } if options == "subtract"
>> > end
>> >
>> > But now I see this output in Rubymonks :
>> >
>> > defaults to addtion when no option is specifiedNoMethodErrorundefined
>> > method
>> > `empty?' for 4:Fixnuminvoking calculate(4, 5, add: true) returns
>> > 9NoMethodErrorundefined method `empty?' for 4:Fixnuminvoking
>> > calculate(-10,
>> > 2, 3, add: true) returns -5NoMethodErrorundefined method `empty?' for
>> > -10:Fixnuminvoking calculate(0, 0, 0, 0, add: true) returns
>> > 0NoMethodErrorundefined method `empty?' for 0:Fixnum
>> >
>> > I think the problem is that I cannot take care that the right values are
>> > in
>> > the right variable.
>> >
>> >
>> > Anyone a tip how to solve this ?
>>
>> The options hash is the first parameter to your function, so pass it
>> as the first parameter:
>>
>> calculate({:add => true}, 4,5)
>>
>> and also, you shouldn't check options == "add", but options[:add] ==
>> true or just options[:add]
>>
>> Jesus.
>>
>
> Thanks,
>
> I can't change the function call because I try to learn ruby by a
> interactive website.
Then you have to modify how you declare the function.
I can declare it as this def calculate (*arguments)
So i will be one hash but then I have to figure out how to seperate the options part and the numbers parts.
back to irb how then arguments will look like and figure it out later.
Roelof
···
On Fri, Sep 28, 2012 at 10:18 AM, Roelof Wobben <rwobben@hotmail.com> wrote:
>> On Fri, Sep 28, 2012 at 9:40 AM, Roelof Wobben <rwobben@hotmail.com> > >> wrote:
>
> I don't understand one remark. You say I shouldn't check the options
> =="add" .
> What I' trying to do is to add the word "add" to options if the options
> array = empty.
> So the function defaults to "add" when nothing is entered.
Options is not an array, it's a hash. but in any case it's never going
to be equal to "add". It will contain an element or a key "add".
I can declare it as this def calculate (*arguments)
So i will be one hash but then I have to figure out how to seperate the
options part and the numbers parts.
Then why do you do it? Just set the parameter list according to the
method call:
def calculate *numbers, options
...
end
I'm also quite sure that you're not supposed to subtract the numbers
from 0 but rather from each other. So calculate(1, 2, {}) should yield 1
- 2 = -1 and not 0 - 1 - 2 = -3.
In this case you must omit the "0" parameter for "inject". And you
should use the short form:
Date: Fri, 28 Sep 2012 18:08:52 +0900
From: lists@ruby-forum.com
Subject: Re: calcaulation with unknown numbers of numbers and options fail
To: ruby-talk@ruby-lang.org
Hi,
Roelof Wobben wrote in post #1077862:
> I can declare it as this def calculate (*arguments)
> So i will be one hash but then I have to figure out how to seperate the
> options part and the numbers parts.
Then why do you do it? Just set the parameter list according to the
method call:
def calculate *numbers, options
...
end
I'm also quite sure that you're not supposed to subtract the numbers
from 0 but rather from each other. So calculate(1, 2, {}) should yield 1
- 2 = -1 and not 0 - 1 - 2 = -3.
In this case you must omit the "0" parameter for "inject". And you
should use the short form:
What I've written down is simply the short form of "inject":
Instead of passing a block, you can also pass a method name or operator
(as a symbol). The elements will then be aggregated using this
method/operator.
So
[1, 2, 3].inject :+
is the same as
[1, 2, 3].inject {|sum, element| sum + element}
But it's obviously shorter. When you use the short form, it's common to
use the method name "reduce" instead of "inject" (but both are the same
method).
And the "operator" variable is used to specify the operator for "reduce"
according to the option hash.
Instead of passing a block, you can also pass a method name or operator
(as a symbol). The elements will then be aggregated using this
method/operator.
So
[1, 2, 3].inject :+
is the same as
[1, 2, 3].inject {|sum, element| sum + element}
But it's obviously shorter. When you use the short form, it's common to
use the method name "reduce" instead of "inject" (but both are the same
method).
And the "operator" variable is used to specify the operator for "reduce"
according to the option hash.
I now finisched Ruby Monks and will go back to ruby koans now.
Roelof
···
Date: Fri, 28 Sep 2012 18:59:50 +0900
From: lists@ruby-forum.com
Subject: Re: calcaulation with unknown numbers of numbers and options fail
To: ruby-talk@ruby-lang.org
Roelof Wobben wrote in post #1077872:
> So if I understand this well then you are saying when options is empty
> or options = add then add the numbers otherwise subtract the numbers ?
Yes. But the subtraction was just a guess. I don't know the actual
assignment.