Why do I get a syntax error using the ternary operator?

Question

I wrote an if-then-else program, but when changing the code to use the ternary operator I am getting a syntax error and cannot figure out why it will not work:

def are_you_over_21
  puts"How old are you?"
  age = gets.chomp.to_i
  age>=21 ? puts"You're over 21!" : puts"You're too young."
end

are_you_over_21

Answer 1

You'll have to think about the Ruby parser.

This will work fine:

age >= 21 ? puts("You're over 21!") : puts("You're too young.")

The reason is that if you use the puts method without brackets, the parser doesn't understand the : token that comes afterwards; it doesn't realise that it is associated with the ternary operator earlier. You can also do this:

puts age >= 21 ? "You're over 21!" : "You're too young."

This is fine -- the parser doesn't run into any ambiguities here.

You can also do this:

puts "You're #{ age >= 21 ? 'over 21!' : 'too young.' }"

Here's why the ambiguity happens. Let's say you have two functions, x and a, like this:

2.0.0-p195 :033 > def x(y)
2.0.0-p195 :034?>   y + 1
2.0.0-p195 :035?>   end
 => nil 
2.0.0-p195 :036 > def a(b)
2.0.0-p195 :037?>   b + 2
2.0.0-p195 :038?>   end
 => nil 

The brackets don't make a difference here:

2.0.0-p195 :039 > x a 1
 => 4 
2.0.0-p195 :040 > x a(1)
 => 4 
2.0.0-p195 :041 > x(a(1))
 => 4 
2.0.0-p195 :042 > x(a 1)
 => 4 

But if x could take more than one parameter? Now you have to help the parser out:

2.0.0-p195 :043 > def x(y, z)
2.0.0-p195 :044?>   y + z + 1
2.0.0-p195 :045?>   end

Let's make a call without brackets like before:

2.0.0-p195 :047 > x a 1, 2 

Are you passing a(1), 2 to x, or are you trying to call a(1, 2) and pass the result to x? By convention Ruby assumes you are trying to call a(1, 2), but a only takes one argument, so you get an error:

ArgumentError: wrong number of arguments (2 for 1)
  from (irb):36:in `a'

So you need brackets:

2.0.0-p195 :052 > x a(1), 2
 => 6