Ruby 脚本忽略条件和中断方法而不会抛出错误

Question

I have this (I believe) straight and easy method meant to verify if a certain string only includes numbers and isn't empty.

class String

    def is_number?
    
        puts "Here it's working, 1"
    
        if self.scan(/\D/).empty? and self != ""
            return true
            puts "true"
        else
            return false
            puts "false"
        end
        
        puts "Here it's working, 2"
    end
    
end

"asd".is_number?
puts "Here it's working, 3"

The result is quite astonishing to me:

The method works until before the conditional. At that point it doesn't go with the "then" nor the "else" options (which, up to today, I never thought to be an option too), and instead breaks the method. Then, it proceeds to the following command. Finally, at the end of the program it sits there without throwing any error.

I honestly don't know how to proceed at this point.

Answer 1

When you used return in a method it will not execute any code after that, if you are expecting true / false to print you should put it above the return statement

def is_number?
  puts "Here it's working, 1"
  if self.scan(/\D/).empty? and self != ""
    puts "true"
    return true
  else
    puts "false"
    return false
  end   
  puts "Here it's working, 2"
end

Note:- "Here it's working, 2" statement will never execute as there will be return statement before that.

Answer 2

it doesn't go with the "then" nor the "else" options

No, this is not what happens here, as described in the answer from Salil.

For the future, if you formulate a hypothesis about your code, you should prove or disprove it. Not for us, for yourself. Else how do you know this is actually what is happening?

For example, something like this would reliably verify that the control does indeed enter one of the conditional branches.

if self.scan(/\D/).empty? and self != ""
  #return true
  #puts "true"
  raise "error from if branch"
else
  #return false
  #puts "false"
  raise "error from else branch"
end