explanation of method

rubyonrails-talk
1

hey i have a method

   def admin_login_required       username, passwd = get_auth_data       self.current_user ||= User.authenticate(username, passwd)

:false if username && passwd

      logged_in? && authorized? ? true : access_denied    end

this is from an authentication plugin. i cant make sense of it, and dont understand wot the '||=' symbol does. i cant google for it, cos google strips the search for the symbol. can someone explain step by step wot the function is doing ?

2

This is a ruby question so probably better suited to other places, but anyway, for most operators a op= b is the same as a = a op b So a ||= b is the same a = a || b Because of the way ruby evaluates these things, it means set a to b unless a is already set (in which case it won't even evaluate b).

Fred

3

Hi --

hey i have a method

  def admin_login_required      username, passwd = get_auth_data      self.current_user ||= User.authenticate(username, passwd) >> :false if username && passwd      logged_in? && authorized? ? true : access_denied   end

this is from an authentication plugin. i cant make sense of it, and dont understand wot the '||=' symbol does. i cant google for it, cos google strips the search for the symbol. can someone explain step by step wot the function is doing ?

This is a ruby question so probably better suited to other places, but anyway, for most operators a op= b is the same as a = a op b So a ||= b is the same a = a || b Because of the way ruby evaluates these things, it means set a to b unless a is already set (in which case it won't even evaluate b).

There's at least one edge-case which reveals that it a ||= b and a = a || b aren't quite the same:

   irb(main):007:0> h = Hash.new(1)    => {}    irb(main):008:0> h[:x] ||= 2    => 1    irb(main):009:0> h    => {}

Here's what happens with the plain || version:

   irb(main):010:0> h[:y] = h[:y] || 3    => 1    irb(main):011:0> h    => {:y=>1}

Matz had something to say about ||= at the Ruby Clinic that I conducted on Friday at RubyConf -- namely, that it's better to think of it this way:

   x ||= y => x || (x = y)

Translating my example that way produces the right result.

David

4

You learn something everyday! Thanks for that.

Fred