What does map(&:name) mean in Ruby?

Question

I found this code in a RailsCast:

def tag_names
  @tag_names || tags.map(&:name).join(' ')
end

What does the (&:name) in map(&:name) mean?

Answer 1

It is same as below:

def tag_names
  if @tag_names
    @tag_names
  else
    tags.map{ |t| t.name }.join(' ')
end

Answer 2

It's equivalent to

def tag_names
  @tag_names || tags.map { |tag| tag.name }.join(' ')
end

Answer 3

It's shorthand for tags.map { |tag| tag.name }.join(' ')

Answer 4

It's shorthand for tags.map(&:name.to_proc).join(' ')

If foo is an object with a to_proc method, then you can pass it to a method as &foo, which will call foo.to_proc and use that as the method's block.

The Symbol#to_proc method was originally added by ActiveSupport but has been integrated into Ruby 1.8.7. This is its implementation:

class Symbol
  def to_proc
    Proc.new do |obj, *args|
      obj.send self, *args
    end
  end
end

Answer 5

Josh Lee's answer is almost correct except that the equivalent Ruby code should have been as follows.

class Symbol
  def to_proc
    Proc.new do |receiver|
      receiver.send self
    end
  end
end

not

class Symbol
  def to_proc
    Proc.new do |obj, *args|
      obj.send self, *args
    end
  end
end

With this code, when print [[1,'a'],[2,'b'],[3,'c']].map(&:first) is executed, Ruby splits the first input [1,'a'] into 1 and 'a' to give obj 1 and args* 'a' to cause an error as Fixnum object 1 does not have the method self (which is :first).

---

When [[1,'a'],[2,'b'],[3,'c']].map(&:first) is executed;

1. :first is a Symbol object, so when &:first is given to a map method as a parameter, Symbol#to_proc is invoked.

2. map sends call message to :first.to_proc with parameter [1,'a'], e.g., :first.to_proc.call([1,'a']) is executed.

3. to_proc procedure in Symbol class sends a send message to an array object ([1,'a']) with parameter (:first), e.g., [1,'a'].send(:first) is executed.

4. iterates over the rest of the elements in [[1,'a'],[2,'b'],[3,'c']] object.

This is the same as executing [[1,'a'],[2,'b'],[3,'c']].map(|e| e.first) expression.

Answer 6

Two things are happening here, and it's important to understand both.

As described in other answers, the Symbol#to_proc method is being called.

But the reason to_proc is being called on the symbol is because it's being passed to map as a block argument. Placing & in front of an argument in a method call causes it to be passed this way. This is true for any Ruby method, not just map with symbols.

def some_method(*args, &block)
  puts "args: #{args.inspect}"
  puts "block: #{block.inspect}"
end

some_method(:whatever)
# args: [:whatever]
# block: nil

some_method(&:whatever)
# args: []
# block: #<Proc:0x007fd23d010da8>

some_method(&"whatever")
# TypeError: wrong argument type String (expected Proc)
# (String doesn't respond to #to_proc)

The Symbol gets converted to a Proc because it's passed in as a block. We can show this by trying to pass a proc to .map without the ampersand:

arr = %w(apple banana)
reverse_upcase = proc { |i| i.reverse.upcase }
reverse_upcase.is_a?(Proc)
=> true

arr.map(reverse_upcase)
# ArgumentError: wrong number of arguments (1 for 0)
# (map expects 0 positional arguments and one block argument)

arr.map(&reverse_upcase)
=> ["ELPPA", "ANANAB"]

Even though it doesn't need to be converted, the method won't know how to use it because it expects a block argument. Passing it with & gives .map the block it expects.