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Ruby Ruby Operators and Control Structures Logical Operators The And (&&) Operator

kori redman
kori redman
4,173 Points
Posted by kori redman
kori redman
4,173 Points

What is wrong with my code? Please help!

Here is what I currently have written.

def check_speed(car_speed) car_speed = get.chomp.to_i safe = (car_speed <= 40) && (car_speed >= 50) return safe end

ruby.rb 
def check_speed(car_speed)
  car_speed = get.chomp.to_i 
  safe = (car_speed <= 40) && (car_speed >= 50) 
  return safe
end

3 Answers

Jason Anders
MOD
Jason Anders
Treehouse Moderator 145,858 Points
Jason Anders
Jason Anders
Treehouse Moderator 145,858 Points

Hi Kori and Welcome to Treehouse.

There are a few things wrong with your code. (1) - I'm not sure why you are using a get.chomp.to_i -- it is not asked for and is not needed. (2) - you have your comparison operators backwards -- you want to check to see if the speed is greater than / equal to 40 and less than / equal to 50, and for this, you need an if statement. (3) - you are returning a variable called safe instead of a string (needs "").

The corrected code will look like this:

def check_speed(car_speed)
  if (car_speed >= 40 && car_speed <= 50)
    return "safe"
  end
end

I hope this makes sense and helps. Keep Coding! :)

kori redman
kori redman
4,173 Points
kori redman
kori redman
4,173 Points

This is sooo helpful!!! Thanks so much!

I was using get.chomp for 2 reasons. 1. It is in the video just previous to this challenge (therefore thought it might secretly be needed and 2. I was trying to go above and beyond.

I be more mindful that Treehouse isn't trying to trick me, in the future! :)

Thanks again!

Jason Anders
Jason Anders
Treehouse Moderator 145,858 Points
Jason Anders
Jason Anders
Treehouse Moderator 145,858 Points

You are very welcome. :)

I hope you enjoy the rest of the Ruby track.

Ps. It is good practice to mark the answer that solved your question with "Best Answer", as this lets others in the community know that your post has been successfully resolved.

Tomasz Jaśkiewicz
Tomasz Jaśkiewicz
5,399 Points
Tomasz Jaśkiewicz
Tomasz Jaśkiewicz
5,399 Points

If I may add something for the future. Look at this methods below:

def check_speed(car_speed)
  case car_speed
  when 40..50
    return :safe
  else
    return :unsafe
  end
end

It will return symbol :safe if speed is in the range [40,50]or will return :unsafe otherwise. What's my point here? There are a few actually:

  1. As you can see all possible cases are analysed here. If speed equals 40 it will return :safe if 20, 51, 60, 100 it will return :unsafe. If you use method called check_speed you expect that it will inform you in all cases whether speed is safe or not. Considering previous method with if statement it will return nil value if speed won't fit in the range. nil is tricky and we shouldn't use it as return value. In this particular case you are checking speed and returning info about its safety. I used case when here. You can simply add next ranges here (eg. define a new range 51..60 and will return :too_high.
  2. I assume you'll be often using this function to check car speeds. You can use symbols instead of strings here. Each string has its own object id and if we have two the same symbols they have the same object id. Take a look here:
[1] pry(main)> "safe".object_id
=> 70324483105860
[2] pry(main)> "safe".object_id
=> 70324490479640
[3] pry(main)> :safe.object_id
=> 2875228
[4] pry(main)> :safe.object_id
=> 2875228

It means that you can simply save some memory :)

  1. If you want only to check if speed is safe you can write much simpler method:
def safe_speed?(car_speed)
  40..50.include?(car_speed)
end

This method will return always boolean value (true or false). Using ? at the end of method name informs us that method will return only boolean values. I'm using everywhere smth like this: 40..50. If you're not familiar with it, it just creates a range from 40 (included) up to 50 (also included).

[1] pry(main)> (40..50).class
=> Range
[2] pry(main)> (40..50).count
=> 11
[3] pry(main)> (40..50).each{ |n| puts n }
40
41
42
43
44
45
46
47
48
49
50
=> 40..50
[4] pry(main)> (40..50).to_a
=> [40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50]

I hope, it will be useful in the future. Have fun with Ruby! :)

kori redman
kori redman
4,173 Points
kori redman
kori redman
4,173 Points

This information is so thorough and I learned some more advanced code (..) to use in the future!

Thanks very much! :) I love coding!