How do I use elsif and ranges? - ruby

Question

I have the following code:
var_comparison = 5
print "Please enter a number: "
my_num = Integer(gets.chomp)
if my_num > var_comparison
print "You picked a number greater than 5!"
elsif my_num < var_comparison
print "You picked a number less than 5!"
elsif my_num > 99
print "Your number is too large, man."
else
print "You picked the number 5!"
end
The interpreter has no way of distinguishing between accepting the rule >5 or >99. How do I make it so that any number between 6-99 returns "You picked a number greater than 5!", but a number 100 or greater returns "Your number is too large, man!"?
Do I need to specifically state a range somehow? How would I best do that? Would it by the normal range methods e.g.
if my_num 6..99
or
if my_num.between(6..99)
?

Answer 1

You can express it as a range, but it would be much simpler to rearrange the order of your conditions to achieve what you want. The interpreter runs through conditional if/else statements in the order they are written, stopping when a condition is true or when else is reached. This makes the order important. We can know that if I get to an elsif, all the preceding conditions must have been false. So in your code:
var_comparison = 5
print "Please enter a number: "
my_num = Integer(gets.chomp)
if my_num > 99
# my_num is > 99
print "Your number is too large, man."
elsif my_num > var_comparison # to get here, my_num must be <= 99
print "You picked a number greater than 5!"
elsif my_num < var_comparison
print "You picked a number less than 5!"
else
print "You picked the number 5!"
end
If you needed to express a number as a range (if your conditional logic becomes more complex), you could do the following:
if (6..99).cover?(my_num)

Answer 2

Clearly the interpreter has no way of distinguishing between accepting the rule >5 or >99.
Yes it does: it tests the conditions in textual order! Since 100 is both greater than 5 and greater than 99, both conditions match, but in an if/elseif chain only one condition is ever evaluated. You should move the clauses around to achieve what you want.

Answer 3

Rather than use elseif, use case/when:
def test_range(foo)
case foo
when 5 .. 99
puts "#{ foo } is in range"
else
puts "#{ foo } is out of range"
end
end
test_range(4)
test_range(5)
test_range(99)
test_range(100)
# >> 4 is out of range
# >> 5 is in range
# >> 99 is in range
# >> 100 is out of range
In your code's case, it needs to be a bit different, because testing for less-than and greater-than doesn't lend itself to the normal range tests, because they expect discrete minimum and maximum values. So, instead, this is what I'd do:
my_num = 5
case
when my_num > 99
puts "Your number is too large, man."
when my_num > 5
puts "You picked a number greater than 5!"
when my_num < 5
puts "You picked a number less than 5!"
else
puts "You picked the number 5!"
end
I find that long chains of if/elseif/else become harder to read, so I prefer case statements.

Answer 4

As Kilian said, if clauses are evaluated in order, so the first to match is considered the chosen one.
You can even use case; when to achieve this:
Infinity = 1.0/0
var_comparison = 5
print "Please enter a number: "
my_num = Integer(gets.chomp)
case my_num
when -Infinity...var_comparison
puts "You picked a number less than #{var_comparison}!"
when var_comparison
puts "You picked the number #{var_comparison}!"
when var_comparison..99
puts "You picked a number greater than #{var_comparison}!"
else
puts "Your number is too large, man."
end