How to Convert a Ruby Proc to Block?

In Ruby, when passing a Proc object to a method as an argument, you can use the unary ampersand operator (&) to convert it to a block, for example, like so:

# create a method that accepts a block
def my_method(&block)
  if block_given?
    # call the block using `yield`
    yield(10)
  else
    "not a block"
  end
end

my_proc = proc { | x | puts x * 2 }

# convert the `Proc` to a block and pass it to the method
puts my_method(&my_proc) #=> 20

This can be useful for passing a Proc (or lambda) to a method that expects a block argument. In the example above, Kernel#block_given? returns true and yield executes the block in the current context as expected.

If you modify the code above and remove the & operator, you will see that the method argument is no longer treated as a block:

def my_method(block)
  if block_given?
    yield(10)
  else
    "not a block"
  end
end

my_proc = proc { | x | puts x * 2 }

puts my_method(my_proc) #=> "not a block"

Please note that when a Proc object is converted to a block using the & operator, it can still be invoked using the Proc#call method as it retains its underlying functionality. This also means that when you call the class method on the object inside the method, it will return "Proc".