lambda's method? Mats' example code confuse me .

Question

def memoize
  cache = {}
  lambda { |*args| 
    unless cache.has_key?(args)
      cache[args] = self[*args]
    end
    cache [args]
  }
end

factorial =  lambda {|x| return 1 if x== 0; x*factorial[x-1];}.memoize

puts factorial.call 10

The code is from book "The ruby programming language ". But it confuse me : how can the method(memoize) apply to a lambda as its method? Can lambda followed by other lambda with dot(.) as its own method?

lambda {|x| return 1 if x== 0; x*factorial[x-1];}.memoize

BTW: The above code works in irb, but ruby interpreter encounter error as following:

memoize.rb:11: private method `memoize' called for #<Proc:[email protected]:11> (NoMethodError)

Why?

Answer 1

Where you're saying this:

def memoize
  #...
end

I think you mean to say this:

class Proc
  def memoize
    #...
  end
end

That would add a public memoize method to Procs and lambda { ... } (or -> { ... } in newer Rubies) gives you a Proc instance.

Now on to memoize itself. Methods return the value of their last expression and for memoize, that last expression is this:

lambda { |*args| 
  unless cache.has_key?(args)
    cache[args] = self[*args]
  end
  cache [args]
}

So memoize returns a wrapper for the Proc (self) that is a closure over cache and all this wrapper does is:

1. Check to see if cache has an entry for the argument list in question (the Array args) .
2. If we don't have a cached value then compute the original Proc's value (self[*args]) and store it in the cache.
3. Return the cached value.

You can use the [] method to execute a Proc so proc.call(a, b) is the same as proc[a, b].