From: "Jesús Gabriel y Galán" <jgabrielygalan@...>
Date: 2009-01-29T01:01:37+09:00
Subject: Re: function inside a function

On Wed, Jan 28, 2009 at 4:49 PM, Jason Lillywhite
<jason.lillywhite@gmail.com> wrote:
> I have a question about the timing of function evaluations in this case:
>
> def function_a(x = 1)
>  x**2 + x
> end
>
> def function_b(any_function)
>  y = 4 + 5 #this would actually be more complex
>  function_a(y)
> end
>
> puts function_b(function_a)
>
> I found that function_a(x = 1) is evaluated first, then
> function_b(function_a). I found that 'def function_a(x = nil) returns an
> error. Why does function_a need to be evaluated before it is called as
> an argument in function_b?

The problem here is that when you write function_a, this is actually
*calling* function_a,which as you are not supplying any parameter uses
the default value of 1 you defined.

> PS - the reason I am writing things this way is to make function_b
> accept any function (within reason).

For sure there are many ways to achieve this, but a common way is to
use blocks and lambdas:

function_a = lambda {|x| x**2 + x}

def function_b
  y = 4 + 5 #this would actually be more complex
  yield y
 end

or more explicitly:

def function_b (&blk)
  y = 4 + 5
  blk.call(y) # or blk[y]
end

puts function_b(&function_a)

Hope this helps,

Jesus.