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Are there more elegant ways to prevent negative numbers in Ruby?

Given that I'd like to do the following calculation:

total = subtotal - discount

Because discount might be greater than subtotal, there is code like the following:

class Calculator
  def initialize(subtotal: subtotal, discount: discount)
    @subtotal = subtotal
    @discount = discount
  end

  def total
    [subtotal - discount, 0].max
  end

  private

  def subtotal
    @subtotal
  end

  def discount
    @discount
  end
end

When seeing the [subtotal - discount, 0].max part or any similar code, I often have to pause and think.

Are there more elegant ways to handle this kind of calculation?

like image 741
Domon Avatar asked Jul 21 '15 04:07

Domon


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2 Answers

I think your solution is essentially correct, and probably the most readable besides a small refactor. I might change it slightly like so:

  def total
    final_total = subtotal - discount
    [final_total, 0].max
  end

The ruby expression [final_total, 0].max is essentially the traditional solution in mathematics for the same function: max {final_total, 0}. The difference is just notation and context. Once you see this max expression once or twice you can read it as follows: "final_total, but at least zero".

Perhaps if you use this expression more than once you can add another at_least_zero method or something like in Shiko's solution.

like image 73
Rex Butler Avatar answered Oct 03 '22 16:10

Rex Butler


Some performance numbers:

                        user     system      total        real
[i, 0.0].max        0.806408   0.001779   0.808187 (  0.810676)
0.0 if i < 0.0      0.643962   0.001077   0.645039 (  0.646368)
0.0 if i.negative?  0.625610   0.001680   0.627290 (  0.629439)

Code:

require 'benchmark'

n = 10_000_000
Benchmark.bm do |benchmark|
  benchmark.report('[value, 0.0].max'.ljust(18)) do
    n.times do |i|
      a = [-1*i, 0.0].max
    end
  end

  benchmark.report('0.0 if value < 0.0'.ljust(18)) do
    n.times do |i|
       a = 0.0 if -1*i < 0.0 
    end
  end

  benchmark.report('0.0 if value.negative?'.ljust(18)) do
    n.times do |i|
      a = 0.0 if (-1*i).negative?
    end
  end
end
like image 24
Nicolás Alonso Pintado Avatar answered Oct 03 '22 16:10

Nicolás Alonso Pintado