Question

Consider the following two peices of ruby code

Example 1

name = user.first_name
round_number = rounds.count
users.each do |u|
  puts "#{name} beat #{u.first_name} in round #{round_number}"
end

Example 2

users.each do |u|
  puts "#{user.first_name} beat #{u.first_name} in #{rounds.count}"
end

For both pieces of code imagine

#user.rb
def first_name
  name.split.first
end

So in a classical analysis of algorithms, the first piece of code would be more efficient, however in most modern compiled languages, modern compilers would optimize the second piece of code to make it look like the first, eliminating the need to optimize code in such maner.

Will ruby optimize or cache values for this code before execution? Should my ruby code look like example 1 or example 2?

Answer 1

Example 1 will run faster, as first_name() is only called once, and it’s value stored in the variable.

In Example 2 Ruby will not memoize this value automatically, since the value could have changed between iterations for the each() loop.

Therefor expensive-to-calculate methods should be explicitly memoized if they are expected to be used more than once without the return value changing.

Making use of Ruby’s Benchmark Module can be useful when making decisions like this. It will likely only be worth memoizing if there are a lot of values in users, or if first_name() is expensive to calculate.