280
votes

For example:

9 / 5  #=> 1

but I expected 1.8. How can I get the correct decimal (non-integer) result? Why is it returning 1 at all?

7
Note that if you're actually using a method to return this value, you don't need to assign it to a variable; simply def method; a - b/8; end would return the result of the calculation from the method, as the last expression in a method call is the return value.Phrogz

7 Answers

289
votes

It’s doing integer division. You can make one of the numbers a Float by adding .0:

9.0 / 5  #=> 1.8
9 / 5.0  #=> 1.8
374
votes

It’s doing integer division. You can use to_f to force things into floating-point mode:

9.to_f / 5  #=> 1.8
9 / 5.to_f  #=> 1.8

This also works if your values are variables instead of literals. Converting one value to a float is sufficient to coerce the whole expression to floating point arithmetic.

180
votes

There is also the Numeric#fdiv method which you can use instead:

9.fdiv(5)  #=> 1.8
39
votes

You can check it with irb:

$ irb
>> 2 / 3
=> 0
>> 2.to_f / 3
=> 0.666666666666667
>> 2 / 3.to_f
=> 0.666666666666667
26
votes

You can include the ruby mathn module.

require 'mathn'

This way, you are going to be able to make the division normally.

1/2              #=> (1/2)
(1/2) ** 3       #=> (1/8)
1/3*3            #=> 1
Math.sin(1/2)    #=> 0.479425538604203

This way, you get exact division (class Rational) until you decide to apply an operation that cannot be expressed as a rational, for example Math.sin.

11
votes

Change the 5 to 5.0. You're getting integer division.

6
votes

Fixnum#to_r is not mentioned here, it was introduced since ruby 1.9. It converts Fixnum into rational form. Below are examples of its uses. This also can give exact division as long as all the numbers used are Fixnum.

 a = 1.to_r  #=> (1/1) 
 a = 10.to_r #=> (10/1) 
 a = a / 3   #=> (10/3) 
 a = a * 3   #=> (10/1) 
 a.to_f      #=> 10.0

Example where a float operated on a rational number coverts the result to float.

a = 5.to_r   #=> (5/1) 
a = a * 5.0  #=> 25.0