CODEWITHSUNDEEP
rubyboolean-logic
Jcode
Jcode

Reputation: 3

Boolean exercise

I am working on the following exercise in a course and I cannot wrap my head around this. I am very new to learning Rails, so please be patient with me. Currently the exercise wants me to write a method named tasty?, which takes one argument: 

def tasty?(ripe)  
end

tasty? should:

Specs are:

describe "tasty?" do  
  it "should return 'Yes' if ripe is true" do  
    expect( tasty?(true) ).to eq("Yes")  
  end  
  it "should return 'Not Yet' if ripe is false" do  
    expect( tasty?(false) ).to eq("Not Yet")  
  end  
end

I wrote this: 

def tasty?(ripe)  
  if "ripe == 'yes'"  
    ( tasty?("true") ).==("yes")  
  end  
  if "ripe == 'not yet'"  
    ( tasty?("false") ).==("not yet")  
  end  
end

And receive this message when I run it: 

exercise.rb:4: warning: string literal in condition  
exercise.rb:7: warning: string literal in condition

Can anyone tell me what I am doing incorrectly? Thank you for your help.

Upvotes: 0

Views: 117

Answers (3)

Cary Swoveland
Cary Swoveland

Reputation: 110685

As @eirikir has shown you the error or your ways, I will address another issue.

Try this:

def tasty?(ripe)
  if ripe==true
    puts "Yes"
  else
    puts "Not Yet"
  end
end

Then

tasty?(true)  #=> "Yes"
tasty?(false) #=> "Not Yet"
tasty?(nil)   #=> "Not Yet"
tasty?(42)    #=> "Not Yet"

The last returns "Not Yet" because:

 42==true #=> false

Now try this:

def tasty?(ripe)
  if ripe
    puts "Yes"
  else
    puts "Not Yet"
  end
end

Then:

tasty?(true)  #=> "Yes"
tasty?(false) #=> "Not Yet"
tasty?(nil)   #=> "Not Yet"
tasty?(42)    #=> "Yes"

tasty?(42) now returns "Yes" because ripe evaluates true. That's because, in a logical expression (e.g., if..) an object evaluates true (is said to be "truthy") if it equals any value other than false or nil. It's said to be "falsy" if it equals false or nil.

@craig has shown how you can simplify the later expression.

Upvotes: 3

craig.kaminsky
craig.kaminsky

Reputation: 5598

Based on the spec, I would write it this way: 

def tasty?(ripe)
  ripe ? 'Yes' : 'Not Yet'
end

This can also be written more verbosely as: 

def tasty?(ripe) 
  if ripe
    'Yes'
  else 
    'Not Yet' 
  end 
end

The parameter to your method, ripe is going to be - based on the spec you provided - true or false. So, in your method you need to check if ripe is true or false and then return the correct string (i.e., Yes or Not yet).

My first example uses the ternary operator which is a shorthand expression for an if/else statement (as shown in my second code block).

Basically, it is just asking if ripe is true and, if so, it returns 'Yes'. If not, it returns 'Not Yet'.

Upvotes: 1

eirikir
eirikir

Reputation: 3842

The error you're receiving is due to a string in the if statement; you should remove the quotes, e.g.:

if ripe == 'yes'

However, since ripe is apparently always a boolean true or false and not a string "yes", you shouldn't be comparing it to "yes". You should be able to pass it directly to the if statement:

if ripe
  …
else
  …
end

Then you can simply return the desired strings for the different conditions:

if ripe
  "Yes"
else
  "Not yet"
end

Does that make sense?

Upvotes: 3

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