using logical operator with this code is saying "unexpected end of input, expecting keyword_end. why?

You need add a "end" tag when your if-statement is finish.

Hi Aaaron, have you completed this challenge yet? I tried it and it passes with what you say you did in your latest comment. Maybe you've done a typo? I wouldn't usually recommend just giving someone an answer but, since you've tried a few things already:

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

Notice lack of 'return'. Return is not needed for this method. Return is used when you want to return a value and exit the method immediately without running any code that is after that statement in the method. Since, if the condition in our 'if' statement is not met, we're happy for the rest of the method to run and return 'nil', we don't need a 'return' statement in this method.

Also, a return value of a method is equal to its last evaluation, so we can just state "safe" and that will be the return value of the method (if the if condition is met). No need for 'return'.

Also, careful with your indentation - the first line of your method in your original post needs to be indented because it is nested within the method.

In your original answer it's worth noting the misuse of gets.chomp.to_i; to use the method, the user must call the method with an argument, and for the sake of the tutorial, we can assume the user knows to pass in a number. So, the method will be called as something like:

check_speed(42)

Therefore, we will already have a value for 'car_speed' - no need to ask the user to input it after the method has been called.

If the method is called without passing in something as an argument, a Ruby error will be raised and the method will not run.

People often prefer one-line syntax for conditionals:

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

Note that no extra 'end' is needed with this syntax. This syntax is not always best - sometimes the long-hand method is better for readability. In this particular case, personally I think this syntax is better, but it's personal preference.

How is this wrong -

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

?

This is the third or fourth time I've been entering code in exactly how I was told to, and the system gives absolutely no clue why it isn't being accepted. Can anyone please tell me what is wrong with this? This is EASILY my biggest pet peeve with treehouse so far, the total lack of explanation when it tells you you're wrong...that really doesn't work in tandem with a system that apparently demands a strict use of syntax.

I have copied and pasted every answer provided in this thread, and none worked.

Yeah, I have done it with the else statement, without it, I've tried it 100 different ways. What I posted was only the simplest, most-stripped down, most recent version. For instance,

def check_speed(car_speed) if (car_speed >= 40) && (car_speed <= 50) puts "safe" else puts "unsafe" end
end

Does not work.

Ok, it's working now. I tried about a dozen different versions with "return" earlier and none of them worked, but who the hell knows. Thank you.