4 ways to clean ranged comparisons in Ruby conditionals

What is the best way to go when we find conditionals that deal with ranges?

fruits = 0
if age > 5 && age < 10
        fruits += 100
elsif age >= 10 && age < 15
    fruits += 200
end     

One part of the problem is that there are way too many ways to express the conditions

# all these conditions are the same
# first
if age > 5 && age < 10
if 5 < age && 10 > age
if age > 5 && 10 < age
if 5 < age && age < 10
# second
elsif age >= 10 && age < 15
elsif age >= 10 && 15 > age
elsif 10 <= age && age < 15
elsif age => 10  && age < 15

A few tips that I find helpful:

1. No greater than >

You can make the boundaries of the range more readable if you limit yourself to the < operator.

fruits = 0
if 5 < age && age < 10 # 5 |---- age is here -----| 10
    fruits += 100
elsif 10 <= age && age < 15 # 10 |---- age is here -----| 15
    fruits += 200
end     

Llewelyn Falco has a nice blog post where he covers this technique in depth.

2. Use Range

Using Ruby ranges are a good way to represent the conditions, and using .. versus ... can act as a substitute for ≥ or ≤

fruits = 0
if (6...10).include?(age) # 6 |---- age is here -----| 10
    fruits += 100
elsif (10...15).include?(age) # 10 |---- age is here -----| 15
    fruits += 200
end     

3. Use case statements

Took me a while to realise this, but when you use case statements, what you are really doing is comparing two things with the threequals operator ===

The cool thing about it is that when you use === with ranges, you are checking for inclusion within the range.

(1..5) === 2

#=> true

Which means we can write something like this

fruits = 0
case age
when 6...10
    fruits += 100
when 10...15
    fruits += 200
end

4. Use #between?

The #between method gives us an idiomatic way to resolve comparisons. Even if it does not work perfectly well for our example due to the mixed presence use of > and ≥ in both conditions, it's good to know we have this in the Ruby toolkit

fruits = 0
case age
if age.between?(6,10)
    fruits += 100
elsif age.between?(10,15)
    fruits += 200
end