How to combine two procs into one?

Question

Just wondering if there's a syntax shortcut for taking two procs and joining them so that output of one is passed to the other, equivalent to:

a = ->(x) { x + 1 }
b = ->(x) { x * 10 }
c = ->(x) { b.( a.( x ) ) }

This would come in handy when working with things like method(:abc).to_proc and :xyz.to_proc

score 8 · Accepted Answer · answered May 28 '13 at 19:32

8

More sugar, not really recommended in production code

class Proc
  def *(other)
    ->(*args) { self[*other[*args]] }
  end
end

a = ->(x){x+1}
b = ->(x){x*10}
c = b*a
c.call(1) #=> 20

answered May 28 '13 at 19:32

Jim Deville

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Very nice! Why wouldn't you recommend it? – Gunchars May 28 '13 at 19:38
Re-opening built-in classes (like Proc), is always risky due to other libraries possibly doing the same, or future changes to Ruby itself. – Jim Deville May 28 '13 at 19:40
2

I'm surprised it's not in the standard library somewhere. Seems like a fairly useful thing. – Gunchars May 28 '13 at 19:44

score 2 · Answer 2 · answered May 28 '13 at 19:26

2

a = Proc.new { |x| x + 1 }
b = Proc.new { |x| x * 10 }
c = Proc.new { |x| b.call(a.call(x)) }

answered May 28 '13 at 19:26

Arman H

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Please explain what sugar is missing from this answer? – Arman H May 28 '13 at 19:28
Sorry if I wasn't clear: you're given `a` and `b`. You need to get `c`. You can of course write a new proc, but I'm wondering if there's a shorter way, like `b.wrap a` or something. – Gunchars May 28 '13 at 19:28

score 2 · Answer 3 · answered May 29 '13 at 06:15

you could create a union operation like so

class Proc
   def union p
      proc {p.call(self.call)}
   end
end
def bind v
   proc { v}
end

then you can use it like this

 a = -> (x) { x + 1 }
 b = -> (x) { x * 10 }
 c = -> (x) {bind(x).union(a).union(b).call}

score 1 · Answer 4 · answered Jun 15 '19 at 09:32

1

An updated answer. Proc composition is already available in Ruby 2.6. There are two methods << and >>, differing in the order of the composition. So now you can do

##ruby2.6
a = ->(x) { x + 1 }
b = ->(x) { x * 10 }
c = a >> b
c.call(1) #=> 20

answered Jun 15 '19 at 09:32

EJAg

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How to combine two procs into one?

4 Answers4

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