# Is this a bug or a whole new concept? **URL:** \ **Category:** Ruby\ **Created:** [March 2, 2010, 3:58am UTC](https://www.ruby-forum.com/t/is-this-a-bug-or-a-whole-new-concept/184412 "2010-03-02T03:58:33Z")\ **Posts on this page:** 3\ **Page:** 1
**Author:** ![scudco](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/aca169/32.png) [@scudco](https://www.ruby-forum.com/u/scudco)\ **Post date:** [March 2, 2010, 3:58am UTC](https://www.ruby-forum.com/t/is-this-a-bug-or-a-whole-new-concept/184412/1 "2010-03-02T03:58:33Z")
Greetings, I got a book on Ruby and I am trying to learn it piece by piece. I was a Perl programmer for a while so the syntax of Ruby is not too hard for me. What I don’t understand is the following and I thank you in advance for your help and input. I have a variable named “num” that is set to “9.” I’m using the num.eql?() method to test whether or not num.eql?(9) and it gives me “true” in IRB - no surprise there. The next one below also gives me “true” since 18/2 is 9. I read in the book that if you want to force the result of an experssion to be a float, make one or both operands a floating number. So we have 18/2.0. But the result is “false.” How come? How come everytime I change either the number to float I get a false even though the expression is correct? At first I thought I’m not dividing numbers - rather maybe I’m dividing the value of the chars rather than their math value. But that’s not true as “18.to\_i/2.to\_i,” just like “18/2,” gives us “true.” I’m all ears to hear the logic behind such behavior Thanks again irb(main):058:0\> num.eql?(9) =\> true irb(main):059:0\> num.eql?(18/2) =\> true irb(main):060:0\> num.eql?(18/2.0) =\> false irb(main):061:0\> num.eql?(18.0/2) =\> false irb(main):062:0\> num.eql?(18.0/2.0) =\> false irb(main):063:0\> num.eql?(18/2) =\> true irb(main):064:0\> num.eql?(18.0/2.to\_f) =\> false irb(main):065:0\> num.eql?(18.to\_f/2.to\_f) =\> false irb(main):066:0\> num.eql?(18.to\_f/2.to\_i) =\> false irb(main):067:0\> num.eql?(18.to\_i/2.to\_i) =\> true irb(main):068:0\> ---
**Author:** ![scudco](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/aca169/32.png) [@scudco](https://www.ruby-forum.com/u/scudco)\ **Post date:** [March 2, 2010, 4:08am UTC](https://www.ruby-forum.com/t/is-this-a-bug-or-a-whole-new-concept/184412/2 "2010-03-02T04:08:47Z")
Apologies…should have checked the documentation first…we are also comparing types with “eql?” ## num.eql?(numeric) =\> true or false Returns true if num and numeric are the same type and have equal values. 1 == 1.0 #=\> true 1.eql?(1.0) #=\> false (1.0).eql?(1.0) #=\> true Adam Aa wrote: > Greetings, > > I got a book on Ruby and I am trying to learn it piece by piece. I was > a Perl programmer for a while so the syntax of Ruby is not too hard for > me. What I don’t understand is the following and I thank you in advance > for your help and input. > > I have a variable named “num” that is set to “9.” I’m using the > num.eql?() method to test whether or not num.eql?(9) and it gives me > “true” in IRB - no surprise there. The next one below also gives me > “true” since 18/2 is 9. I read in the book that if you want to force > the result of an experssion to be a float, make one or both operands a > floating number. So we have 18/2.0. But the result is “false.” How come? > How come everytime I change either the number to float I get a false > even though the expression is correct? At first I thought I’m not > dividing numbers - rather maybe I’m dividing the value of the chars > rather than their math value. But that’s not true as “18.to\_i/2.to\_i,” > just like “18/2,” gives us “true.” > > I’m all ears to hear the logic behind such behavior > > Thanks again > > irb(main):058:0\> num.eql?(9) > =\> true > irb(main):059:0\> num.eql?(18/2) > =\> true > irb(main):060:0\> num.eql?(18/2.0) > =\> false > irb(main):061:0\> num.eql?(18.0/2) > =\> false > irb(main):062:0\> num.eql?(18.0/2.0) > =\> false > irb(main):063:0\> num.eql?(18/2) > =\> true > irb(main):064:0\> num.eql?(18.0/2.to\_f) > =\> false > irb(main):065:0\> num.eql?(18.to\_f/2.to\_f) > =\> false > irb(main):066:0\> num.eql?(18.to\_f/2.to\_i) > =\> false > irb(main):067:0\> num.eql?(18.to\_i/2.to\_i) > =\> true > irb(main):068:0\> ---
**Author:** ![scudco](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/aca169/32.png) [@scudco](https://www.ruby-forum.com/u/scudco)\ **Post date:** [March 2, 2010, 4:09am UTC](https://www.ruby-forum.com/t/is-this-a-bug-or-a-whole-new-concept/184412/3 "2010-03-02T04:09:47Z")
On Mon, Mar 1, 2010 at 8:59 PM, Adam Aa [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote: > “true” in IRB - no surprise there. The next one below also gives me > “true” since 18/2 is 9. I read in the book that if you want to force > the result of an experssion to be a float, make one or both operands a > floating number. So we have 18/2.0. But the result is “false.” How come? > How come everytime I change either the number to float I get a false > even though the expression is correct? At first I thought I’m not > dividing numbers - rather maybe I’m dividing the value of the chars > rather than their math value. But that’s not true as “18.to\_i/2.to\_i,” > just like “18/2,” gives us “true.” > > I’m all ears to hear the logic behind such behavior #eql? checks to see if the receiver and the parameter have the same value. In Ruby, an integer like 9 is guaranteed not to be identical to a float like 9.0, even if mathematically they are the same (and the == method says they are equal). The result of floating-point division (i.e. where one or both operands is a float) is a floating-point number. From [http://ruby-doc.org/core/classes/Object.html#M000347](http://ruby-doc.org/core/classes/Object.html#M000347): Equality—At the Object level, == returns true only if obj and other are the same object. Typically, this method is overridden in descendent classes to provide class-specific meaning. Unlike ==, the equal? method should never be overridden by subclasses: it is used to determine object identity (that is, a.equal?(b) iff a is the same object as b). The eql? method returns true if obj and anObject have the same value. Used by Hash to test members for equality. For objects of class Object, eql? is synonymous with ==. Subclasses normally continue this tradition, but there are exceptions. Numeric types, for example, perform type conversion across ==, but not across eql?, so: 1 == 1.0 #=\> true 1.eql? 1.0 #=\> false