Ruby - What is this output

I know that this code may be not quite correct:

def print_string(&str)
puts str
end

print_string{"Abder-Rahman"}

But, when I run it, this is what I get:

#<Proc:0x03e25开发者_Go百科d98@r.rb:5>

What is this output?

That's the default string representation of a Proc object. Because "Abder-Rahman" is in braces, Ruby thinks you're defining a block. Did you mean to put str.call inside of your function definition? That should call your block and return the string expression you defined inside it.

The problem is that you've declared that the "print_string" method takes a block argument (confusingly named "str") and you simply print the proc itself. You'd probably like to call the given procedure to see the string value it returns:

def call_proc(&proc)
  proc.call
end
call_proc { 'Foobar' }
# => "Foobar"

What you've discovered is the syntax sugar that if you decorate the last argument of a method definition with an ampersand & then it will be bound to the block argument to the method call. An alternative way of accomplishing the same task is as follows:

def call_proc2
  yield if block_given?
end
call_proc2 { 'Example' }
# => 'Example'

Note also that procedures can be handled directly as objects by using Proc objects (or the "lambda" alias for the Proc constructor):

p1 = Proc.new { 'Foo' }
p1.call # => "Foo"
p2 = lambda { 'Bar' }
p2.call # => "Bar"

You're passing a block to the method, as denoted by the & prefix and how you're calling it. That block is then converted into a Proc internally.

puts str.call inside your method would print the string, although why you'd want to define the method this way is another matter.

See Proc: http://www.ruby-doc.org/core/classes/Proc.html

When the last argument of function/method is preceded by the & character, ruby expect a proc object. So that's why puts's output is what it is.

This blog has an article about the unary & operator.