Ruby：将`＆proc`和`proc`传递给方法之间的区别

Question

I have a bit of code modified slightly from a question on codeacademy. The code is :

def print_list(array, first = 1)
  counter = first
  array.each do |array|
    puts "#{yield counter} #{array}"
    counter = counter.next
  end
end

proc = Proc.new do |n|
  "[#{100*n}]:"
end

print_list ["alpha", "beta", "gamma"], 5, &proc

If I remove the & from the last line Ruby throws me an argument error. What is the purpose of the & here?

Answer 1

If you remove the & , then print_list treats proc as a third argument to itself, instead of a block. The & symbol transforms the Proc object into a block, which is called inside print_list by the yield keyword.

More succinctly, proc is just an argument, &proc is a reference to the block passed to the method.

You might find this article useful to understand the differences between proc and blocks

Answer 2

The & indicates that the proc should be passed as a block.

Without it, the "proc" will just be another (third) parameter so you'll get the argument error (3 for 2)

It's possible to pass it without the & and use it in your print_list method directly as a proc... but first can't be optional then. You'll need to pass first or at the very least nil.

def print_list(array, first, proc)
  counter = first || 1
  array.each do |array|
    puts "#{proc.call counter} #{array}"
    counter = counter.next
  end
end