问题
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?
回答1:
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
回答2:
Another cool shorthand, unknown to many, is
array.each(&method(:foo))
which is a shorthand for
array.each { |element| foo(element) }
By calling method(:foo) we took a Method object from self that represents its foo method, and used the & to signify that it has a to_proc method that converts it into a Proc.
This is very useful when you want to do things point-free style. An example is to check if there is any string in an array that is equal to the string "foo". There is the conventional way:
["bar", "baz", "foo"].any? { |str| str == "foo" }
And there is the point-free way:
["bar", "baz", "foo"].any?(&"foo".method(:==))
The preferred way should be the most readable one.
回答3:
It's equivalent to
def tag_names
@tag_names || tags.map { |tag| tag.name }.join(' ')
end
回答4:
While let us also note that ampersand #to_proc magic can work with any class, not just Symbol. Many Rubyists choose to define #to_proc on Array class:
class Array
def to_proc
proc { |receiver| receiver.send *self }
end
end
# And then...
[ 'Hello', 'Goodbye' ].map &[ :+, ' world!' ]
#=> ["Hello world!", "Goodbye world!"]
Ampersand & works by sending to_proc message on its operand, which, in the above code, is of Array class. And since I defined #to_proc method on Array, the line becomes:
[ 'Hello', 'Goodbye' ].map { |receiver| receiver.send( :+, ' world!' ) }
回答5:
It's shorthand for tags.map { |tag| tag.name }.join(' ')
回答6:
tags.map(&:name)
is The same as
tags.map{|tag| tag.name}
&:name just uses the symbol as the method name to be called.
回答7:
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;
:firstis a Symbol object, so when&:firstis given to a map method as a parameter, Symbol#to_proc is invoked.map sends call message to :first.to_proc with parameter
[1,'a'], e.g.,:first.to_proc.call([1,'a'])is executed.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.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.
回答8:
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.
回答9:
(&:name) is short for (&:name.to_proc) it is same as tags.map{ |t| t.name }.join(' ')
to_proc is actually implemented in C
回答10:
Although we have great answers already, looking through a perspective of a beginner I'd like to add the additional information:
What does map(&:name) mean in Ruby?
This means, that you are passing another method as parameter to the map function. (In reality you're passing a symbol that gets converted into a proc. But this isn't that important in this particular case).
What is important is that you have a method named name that will be used by the map method as an argument instead of the traditional block style.
回答11:
map(&:name) takes an enumerable object (tags in your case) and runs the name method for each element/tag, outputting each returned value from the method.
It is a shorthand for
array.map { |element| element.name }
which returns the array of element(tag) names
回答12:
Here :name is the symbol which point to the method name of tag object.
When we pass &:name to map, it will treat name as a proc object.
For short, tags.map(&:name) acts as:
tags.map do |tag|
tag.name
end
回答13:
it means
array.each(&:to_sym.to_proc)
回答14:
It basically execute the method call tag.name on each tags in the array.
It is a simplified ruby shorthand.
回答15:
It is same as below:
def tag_names
if @tag_names
@tag_names
else
tags.map{ |t| t.name }.join(' ')
end
来源：https://stackoverflow.com/questions/25265311/what-is-the-use-of-the-operator-in-ruby