Tutorial :How can two arguments be passed to a method with a one-argument signature?


s = Proc.new {|x|x*2}    def one_arg(x)    puts yield(x)  end    one_arg(5, &s)  

How does one_arg know about &s?


The & operator turns the Proc into a block, so it becomes a one-argument method with a block (which is called with yield). If you had left off the & so that it passed the Proc directly, you would have gotten an error.


By doing the &s, you're telling one_arg that you'd like your Proc s passed as a block (please correct me if I'm wrong). An equivalent writing would be

one_arg(5) do |x|    x *2  end  

There have been a few questions here on SO as of late that deal with this. August Lilleaas has a pretty nice write up about some of the intricacies of all this Ruby madness.

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