If you think about it there is only one possible answer to that question. How could it evaluate meth2(meth3) without evaluating meth3 first, in order to pass the result to meth2? Similarly how could it call meth1 before it evaluated the parameter to pass to it?
Colin On 23 December 2012 08:14, 7stud -- <li...@ruby-forum.com> wrote: > In this method call: > > meth1(meth2(meth3)) > > ...which value has has to be computed first so that meth1 can return? > > -- > Posted via http://www.ruby-forum.com/. > > -- > You received this message because you are subscribed to the Google Groups > "Ruby on Rails: Talk" group. > To post to this group, send email to rubyonrails-talk@googlegroups.com. > To unsubscribe from this group, send email to > rubyonrails-talk+unsubscr...@googlegroups.com. > For more options, visit https://groups.google.com/groups/opt_out. > > -- You received this message because you are subscribed to the Google Groups "Ruby on Rails: Talk" group. To post to this group, send email to rubyonrails-talk@googlegroups.com. To unsubscribe from this group, send email to rubyonrails-talk+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/groups/opt_out.
