Passing procs and methods

Question

I came across this code:

squareIt = Proc.new do |x| x * x end doubleIt = Proc.new do |x| x + x end def compose proc1, proc2 Proc.new do |x| proc2.call(proc1.call(x)) end end doubleThenSquare = compose(doubleIt, squareIt) squareThenDouble = compose(squareIt, doubleIt) doubleThenSquare.call(5) squareThenDouble.call(5)

doubleThenSquare is called with 5. doubleThenSquare is equal to the return value of compose, which has its two parameters doubleIt and squareIt passed.

I don't see how 5 is passed all its way into the different procs Proc.new do |x|. How does it know what x is in each case?

Cary Swoveland · Accepted Answer · 2018-04-16 21:42:42Z

3

Let's step through it.

doubleIt = Proc.new do |x| x + x end #=> #<Proc:0x00000002326e08@(irb):1429> squareIt = Proc.new do |x| x * x end #=> #<Proc:0x00000002928bf8@(irb):1433> proc1 = doubleIt proc2 = squareIt

compose returns the proc proc3.

proc3 = Proc.new do |x| proc2.call(proc1.call(x)) end #=> #<Proc:0x000000028e7608@(irb):1445>

proc3.call(x) is executed in the same way as

proc3_method(x)

where

def proc3_method(x) y = proc1.call(x) proc2.call(y) end

When x = 5,

y = proc1.call(5) #=> 10 proc2.call(10) #=> 100

proc3_method(5) therefore returns 100, as does proc3.call(5).

answered Apr 16, 2018 at 21:42

Cary Swoveland

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5 Comments

msmith1114 Over a year ago

This step through makes sense, the part thats difficult for me is how the (5) is making it's way to "proc3" in the original code. I know we "call" doubleThenSquare but doubleThenSquare itself isn't a proc function, it calls a method with a proc function in it (lets call that proc3). So thats where my confusion lies. Since we are actually calling something that is storing the return value of a method WITH a proc function in it...but not an actual proc function itself.

Cary Swoveland Over a year ago

doubleThenSquare is a variable that is assigned the proc returned by compose(doubleIt, squareIt). That proc has a variable, but that variable is not assigned a value (e.g. 5) until doubleThenSquare is called.

msmith1114 Over a year ago

Ah so it's able to pass it because a proc is being returned by compose...man that's confusing, since I assumed calling a proc it would need a variable directly (whatever you are calling)

Cary Swoveland Over a year ago

Yes, you've got it. There's an analogous situation with methods. Suppose def m(x) 2*x end. We could then write method(:m).call(3) #=> 6 or public_send(:m, 3) #=> 6. method(:m) converts a method name (:m) to a method object. Being an object, it can be passed around like a proc, and like a proc, is invoked with call (but with Method#call, not Proc#call).

msmith1114 Over a year ago

I guess it's not too crazy hopefully for that to be slighty confusing to me I hope haha (I just started learning ruby)

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