CODEWITHSUNDEEP
ruby
Jacob Crofts
Jacob Crofts

Reputation: 186

Ruby switch statement always evaluates to default

If I simulate the rolling of a six-sided die as follows, it always evaluates to 6. How do I fix this?

def rolldice()
  roll = rand()
  case roll
    when 0..(1/6) then return 1
    when (1/6)..(2/6) then return 2
    when (2/6)..(3/6) then return 3
    when (3/6)..(4/6) then return 4
    when (4/6)..(5/6) then return 5
    else return 6
  end
end

die1 = rolldice()
puts die1

Upvotes: 1

Views: 77

Answers (4)

Dave Sexton
Dave Sexton

Reputation: 11188

When you divide two integers then the result will be an integer so your code is actually doing this:

when 0..0 then return 1
when 0..0 then return 2
when 0..0 then return 3
when 0..0 then return 4
when 0..0 then return 5
else return 6

Try this simple example in IRB:

1 / 6
=> 0

Upvotes: 1

Pedro Ivan
Pedro Ivan

Reputation: 332

Case statement is not the best option to you. It may fit best for texts or fix values. Assuming that you're doing an RPG application, the Switch may be more useful to choose what to do with weapons/armors.

Case is a more elegant way to write a lot of if elses, but behave almost the same way. Before writing the Case you should test the conditions using irb (or rails console if you prefer). This make things easier.

Case also do a lot of comparisons, in this particular case they are not necessary, using the rand() directly will perform better and the code will be simpler. Note that rand(6) will return from 0 to 5, just sum 1 to get the d6.

def rolldice
   return rand(6)+1
end

Upvotes: 0

David Ljung Madison Stellar
David Ljung Madison Stellar

Reputation: 3225

Stefan is right but the answer doesn't clearly explain your problem.

The problem is that rand() gives you a Float (which can have numbers after the decimal point).

But in the case statement you are using ranges such as 0..(1/6).

The problem there is that the numbers in (1/6) are both integers, so the answer is an integer. That's why (1/6) is 0, it's rounded.

You could also solve the problem by using when ranges such as:

when (3.0/6)..(4.0/6) then return 4

Looking at (3.0/6) we have a Float divided by an Integer, the result is a Float, in this case 0.5.

But in the end it's probably best to follow Stefan's suggestion. Instead of using rand for a Float from 0 to 1, we can use a range. Some options:

rand(1..6)    # Returns an Integer from 1 to 6
rand(6)   # Returns an Integer from 0 to 5

If you use the first case, then you can rewrite the when as just when 5... or so on. In this case, because all you want is to return the number, you don't even need the case statement, hence Stefan's shorter answer of just calling rand directly and using that value.

Upvotes: 1

Stefan
Stefan

Reputation: 114218

This is because 1/6 is 0. You can simply pass a range to rand:

def rolldice
  rand(1..6)
end

Upvotes: 5

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