Number of parameters based on variable

Question

Is there a way have x amount of same parameters in Ruby?

Easiest way to ask this is, can you shorten this?

arr = [0,1,2,3]
if x == 1
    return arr
elsif x == 2
    return arr.product(arr)
elsif x == 3
    return arr.product(arr, arr)
elsif x == 4
    return arr.product(arr, arr, arr)
elsif x == 5
    return arr.product(arr, arr, arr, arr)
end

The most essential concept here is that you can use a "splat" (the asterisk *) to turn an array into sequential arguments. The posted answer includes a good example of this. — max pleaner
– max pleaner, Commented Aug 16, 2016 at 4:37
If any of the answers were helpful, please consider selecting the one you preferred. (I'd prefer saying this if there were more than one answer, but the question is several days old...) — Cary Swoveland
– Cary Swoveland, Commented Aug 20, 2016 at 22:51

Cary Swoveland · Accepted Answer · 2016-08-16 04:32:01Z

8

You can obtain the desired result as follows.

def prod(arr, x)
  return arr if x==1
  arr.product(*[arr]*(x-1))
end

arr = [0,1,2,3]

arr                             == prod(arr, 1) #=> true
arr.product(arr)                == prod(arr, 2) #=> true
arr.product(arr, arr)           == prod(arr, 3) #=> true
arr.product(arr, arr, arr)      == prod(arr, 4) #=> true
arr.product(arr, arr, arr, arr) == prod(arr, 5) #=> true

edited Aug 16, 2016 at 4:32

answered Aug 16, 2016 at 4:05

Cary Swoveland

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2 Comments

Santhosh Over a year ago

A one liner return x==1 ? arr : arr.product(*[arr]*(x-1)) :)

Stefan Over a year ago

I'd probably use the array constructor to avoid confusion between the two *'s, i.e. arr.product(*Array.new(x - 1, arr))

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