Ruby：计数器，计数和递增

Question

If you have seen my previous questions, you'd already know I am a big nuby when it comes to Ruby. So, I discovered this website which is intended for C programming, but I thought whatever one can do in C, must be possible in Ruby (and more readable too).

The challenge is to print out a bunch of numbers. I discovered this nifty method .upto() and I used a block (and actually understanding its purpose). However, in IRb, I got some unexpected behavior.

class MyCounter
    def run 
    1.upto(10) { |x| print x.to_s + " " } 
    end
end


irb(main):033:0> q = MyCounter.new
=> #<MyCounter:0x5dca0>
irb(main):034:0> q.run
1 2 3 4 5 6 7 8 9 10 => 1

I have no idea where the => 1 comes from :S Should I do this otherwise? I am expecting to have this result:

1 2 3 4 5 6 7 8 9 10

Thank you for your answers, comments and feedback!

Answer 1

The "=> 1" is from IRB, not your code. After every statement you type into IRB, it prints the result of that statement after a "=>" prompt.

Try printing a newline in your function:

def run 
  1.upto(10) { |x| print x.to_s + " " }
  print "\n"
end

Then it'll look like this:

irb> q.run
1 2 3 4 5 6 7 8 9 10
  => nil

Answer 2

I have no idea where the => 1 comes from

Don't worry. By default irb prints the returning value of the execution of the method.

Even if you don't write the return statement ( like in C for instance ) Ruby returns the value of the last computed statement.

In this case it was 1

That's all.

For instance try:

class WhereIsTheReturn
    def uh?
        14 * 3 # no return keyword
    end
end


whereIsIt = WhereIsTheReturn.new
hereItIs = whereIsIt.uh?
print "Here it is : #{hereItIs}\n"