This code works as expected:
if phrase.last.eql? "?" ? true : false true else false end but this code using the Ruby ternary operator:
phrase.last.eql? "?" ? true : false gives the following error:
warning: string literal in condition
Do I need to escape the "?" somehow?
Without parentheses, ruby is interpreting it as
phrase.last.eql?( "?" ? true : false ) which explains the message "warning: string literal in condition".
To fix this, use parentheses on the parameter:
phrase.last.eql?("?") ? true : false Of course, in this case using the ternary operator is redundant since this is the same as simply
phrase.last.eql?("?") write as below :
phrase.last.eql?("?") ? true : false Example :
2.0.0-p0 :023 > x = "a" => "a" 2.0.0-p0 :024 > x.eql? "?" ? 1 : 2 (irb):24: warning: string literal in condition => false 2.0.0-p0 :025 > x.eql?("?") ? 1 : 2 => 2 2.0.0-p0 :026 > Otherwise x.eql? "?" ? 1 : 2 is interpreted as x.eql?("?" ? 1 : 2). Now in Ruby except nil and false all objects are true. Thus here "?" ? 1 : 2, "?" always will be true, so you got a warning for it. Putting a ever truth value in the conditional test, has no meaning, and the same warning is being thrown to you fro Ruby interpreter.
puts "#{5 == 5.0}, #{5.eql?(5.0)}, #{5.equal?(5.0)}" # => true false false.truebecause the values are the same,false #1because the values are different types, even though they are (=) equal,false #2because the values have differentobject_id's.