Why would one use the unary operator on a property in ruby? i.e &:first

Question

Possible Duplicate:
Ruby/Ruby on Rails ampersand colon shortcut

As a habit I try and read a little of someone elses source code regularly and comment on it in a gist. Right now I'm reading through sinatra's base app and came upon an interesting bit of code (this is part of their Request class)

def accept
  @env['sinatra.accept'] ||= begin 
    entries = @env['HTTP_ACCEPT'].to_s.split(',') 
    entries.map { |e| accept_entry(e) }.sort_by(&:last).map(&:first)
  end
end

The part I don't get is what is &:last and &:first doing?!? It appears as madness!

closer to http://stackoverflow.com/questions/1217088/what-does-mapname-mean-in-ruby — Simon, Jun 09 '11 at 07:00

score 1 · Accepted Answer · answered Jun 09 '11 at 07:08

Read the answers in the duplicate questions for the meaning and usage of &:.... In this case, entries is an array, and there are three methods map, sort_by, and map chained. sort_by(&:last) is equivalent to sort_by{|x| x.last}. map(&:first) is the same as map{|x| x.first}. The reason the first map does not use &:... is because (i) the receiver of accept_entry is not e, and (ii) it takes an argument e.

Why would one use the unary operator on a property in ruby? i.e &:first

1 Answers1

Linked