Solution:Step 1: Understand each Proc's effectp1 converts number to string, p2 adds "!", p3 reverses string.Step 2: Find composition order to get "5!""5".reverse = "5", so p1 >> p3 >> p2 gives: "5" -> "5" -> "5!". But p1 >> p2 >> p3: "5" -> "5!" -> "!5". Only A produces "5!".Final Answer:combined =

[Solved] Given these Procs:p1 = Proc.new { |x| x.to_s } p2 = Proc.new { |s| s + "!" } p3 = Proc.new { |str| str.reverse }How do you compose them so that calling the combined Proc with 5 returns "5!"? | Ruby

Ruby - Functional Patterns in Ruby

Given these Procs:

p1 = Proc.new { |x| x.to_s }
p2 = Proc.new { |s| s + "!" }
p3 = Proc.new { |str| str.reverse }

How do you compose them so that calling the combined Proc with 5 returns "5!"?

Acombined = p3 >> p2 >> p1

Bcombined = p1 >> p2 >> p3

Ccombined = p1 >> p3 >> p2

Dcombined = p2 >> p1 >> p3

Step-by-Step Solution

Solution:

Step 1: Understand each Proc's effect
p1 converts number to string, p2 adds "!", p3 reverses string.
Step 2: Find composition order to get "5!"
"5".reverse = "5", so p1 >> p3 >> p2 gives: "5" -> "5" -> "5!". But p1 >> p2 >> p3: "5" -> "5!" -> "!5". Only A produces "5!".
Final Answer:
combined = p1 >> p3 >> p2 -> Option C
Quick Check:
to_s, reverse, then add ! = "5!" [OK]

Quick Trick: Order Procs by desired transformations stepwise [OK]

Common Mistakes:

Putting p2 before p3 changes output to "!5"
Reversing order causes errors on non-string
Ignoring effect of reverse on "5!"

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Given these Procs:

Step 1: Understand each Proc's effect

Step 2: Find composition order to get "5!"

Final Answer:

Quick Check:

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