What do you call the &: operator in Ruby? [duplicate]

ghz 1years ago ⋅ 5606 views

Question

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Closed 13 years ago.

Possible Duplicates:
[Ruby/Ruby on Rails ampersand colon shortcut](https://stackoverflow.com/questions/1961030/ruby-ruby-on-rails- ampersand-colon-shortcut)
[What does map(&:name) mean in Ruby?](https://stackoverflow.com/questions/1217088/what-does-mapname-mean- in-ruby)

I was reading Stackoverflow and stumbled upon the following code

array.map(&:to_i)

Ok, it's easy to see what this code does but I'd like to know more about &: construct which I have never seen before.

Unfortunately all I can think of is "lambda" which it is not. Google tells me that lambda syntax in Ruby is ->->(x,y){ x * y }

So anyone knows what that mysterious &: is and what it can do except calling a single method?


Answer

There's a few moving pieces here, but the name for what's going on is the [Symbol#to_proc](http://ruby-doc.org/core-1.9.3/Symbol.html#method-i- to_proc) conversion. This is part of Ruby 1.9 and up, and is also available if you use later-ish versions of Rails.

First, in Ruby, :foo means "the symbol foo", so it's actually two separate operators you're looking at, not one big &: operator.

When you say foo.map(&bar), you're telling Ruby, "send a message to the foo object to invoke the map method, with a block I already defined called bar". If bar is not already a Proc object, Ruby will try to make it one.

Here, we don't actually pass a block, but instead a symbol called bar. Because we have an implicit to_proc conversion available on Symbol, Ruby sees that and uses it. It turns out that this conversion looks like this:

def to_proc
  proc { |obj, *args| obj.send(self, *args) }
end

This makes a proc which invokes the method with the same name as the symbol. Putting it all together, using your original example:

array.map(&:to_i)

This invokes .map on array, and for each element in the array, returns the result of calling to_i on that element.