# Objective

Learn what happens when you don't have an algorithm to guide your implementation of the solution.

# Problem Statement

Convert the character representation of an integer to it's decimal format.

# Discussion

Computers can store data in bits (zeroes and ones) which are numbers that can be converted to decimal, octal etc. They cannot store letters or other special symbols. ASCII character encoding allows computers to store letters, text, symbols and control characters.

# Solution Domain Analysis

Let's convert 4 character sequence '1984' to the decimal number 1984.

Here is the ASCII table, you can see the decimal representation of the characters '0' to '9'.

To convert, 49 needs to be converted to 1000, 57 to 900, 56 to 80 and 52 to 4 units. To get the decimal digit we have to subtract 48 (ASCII value of character '0') from each of the decimal value for the character.

To convert the one-character string to it's decimal representation in ASCII we can use the ord method of String class in Ruby. Playing in the irb:

``````  > '0'.ord
=> 48
> '1'.ord
=> 49
> '9'.ord
=> 57
> '8'.ord
=> 56
> '4'.ord
=> 52
``````

First number = 49 - 48 = 1
Second number = 57 - 48 = 9
Third number = 56 - 48 = 8
Fourth number = 52 - 48 = 4

The shifting to the left mechanism can be obtained at each step by multiplying the previous decimal value by 10 and adding it to current decimal digit.

# Test Cases

``````class CharacterConverter

end

describe CharacterConverter do
it "should convert '0' to 0"
it "should convert '1' to 1"
it "should convert '10' to 10"
it "should convert '100' to 100"
it "should convert '1000' to 1000"
end
``````

Here I have listed the tests in increasing level of difficulty. I have also chosen the data set that is lowest to test that the solution works.

# Steps

## Step 1

``````class CharacterConverter
def initialize(n)
@n = n
end

def to_i
@n.to_i
end
end

describe CharacterConverter do
it "should convert '0' to 0" do
cc = CharacterConverter.new('0')
result = cc.to_i
expect(result).to eq(0)
end

it "should convert '1' to 1"
it "should convert '10' to 10"
it "should convert '100' to 100"
it "should convert '1000' to 1000"
end
``````

Simplest implementation passes.

## Step 2

Let's tackle the second test.

``````  it "should convert '1' to 1" do
cc = CharacterConverter.new('1')
result = cc.to_i
expect(result).to eq(1)
end
``````

This passes without any change to the code.

## Step 3

``````  it "should convert '1984' to 1984" do
cc = CharacterConverter.new('1984')
result = cc.to_i
expect(result).to eq(1984)
end
``````

Also passes. We don't want to use Ruby's `to_i` method which does the conversion. We want to develop our own implementation of `to_i`.

## Step 4

Let's not call the `to_i` on the String.

``````  def to_i
'0'.ord - @n.ord
end
``````

This passes first test.

## Step 5

``````  it "should convert '10' to 10" do
cc = CharacterConverter.new('10')
result = cc.to_i
expect(result).to eq(10)
end
``````

It fails. There is gap in our problem domain analysis. We have not thought about how many digits there are in a given string, so we don't know when to terminate. Reading the Ruby documentation, I found a method that solves that problem for us. Playing in the irb:

`````` > n = '1984'
> n.bytes
=> [49, 57, 56, 52]
``````

## Step 6

``````class CharacterConverter
def initialize(n)
@n = n
@numbers = n.bytes
end

def to_i
first_element = @numbers.shift
first_number = (first_element.ord - '0'.ord)

if @numbers.size > 0
next_element = @numbers.shift
next_number = (next_element.ord - '0'.ord)
puts 'inside if'
first_number * 10 + next_number
else
puts 'hi'
first_number
end
end
end

describe CharacterConverter do
it "should convert '0' to 0" do
cc = CharacterConverter.new('0')
result = cc.to_i
expect(result).to eq(0)
end

it "should convert '1' to 1" do
cc = CharacterConverter.new('1')
result = cc.to_i
expect(result).to eq(1)
end

it "should convert '10' to 10" do
cc = CharacterConverter.new('10')
result = cc.to_i
expect(result).to eq(10)
end

xit "should convert '1984' to 1984" do
cc = CharacterConverter.new('1984')
result = cc.to_i
expect(result).to eq(1984)
end

end
``````

This implementation passes all three tests. I had to add puts statements to evolve the logic to get the tests passing. The puts statement is the simplest debugging tool. Why do we need debugger when we are using TDD? The test was forcing the code to evolve fast so I had to see what was happening. Adding assertion to check our code is another alternative. But the implementation details cannot be exposed to the test, so I used the print statements. This happened because that I did not have an algorithm as a guide for my code. I had to come up with the solution by trial and error.

## Step 7

``````  it "should convert '1984' to 1984" do
cc = CharacterConverter.new('1984')
result = cc.to_i
expect(result).to eq(1984)
end
``````

It fails.

## Step 8

Let's take a step back. Make the failing test pending for now. Let's refactor to get the code ready to be generalized. After cleanup:

``````  def to_i
first_element = @numbers.shift
first_number = (first_element.ord - '0'.ord)
result = first_number

if @numbers.size > 0
next_element = @numbers.shift
next_number = (next_element.ord - '0'.ord)
result = first_number * 10 + next_number
end
result
end
``````

## Step 9

Run the story test now. It will fail. Change the implementation.

``````  def to_i
first_element = @numbers.shift
first_number = (first_element.ord - '0'.ord)

previous_number = first_number
while @numbers.size > 0
next_element = @numbers.shift
next_number = (next_element.ord - '0'.ord)
previous_number = previous_number * 10 + next_number
end
previous_number
end
``````

The story test passes. This is a 1-2-3 punch, `BOOM` here is the solution example. In order to arrive at this solution, you must know how to apply reduction process, terminating condition and initial condition that we have discussed in previous articles.

## Step 10

Let's cleanup.

``````  def to_i
first_element = @numbers.shift
first_number = ascii_value(first_element)

previous_number = first_number
while not_complete?
next_element = @numbers.shift
next_number = ascii_value(next_element)
previous_number = previous_number * 10 + next_number
end
previous_number
end

private

def ascii_value(n)
n.ord - '0'.ord
end

def not_complete?
@numbers.size > 0
end
``````

Idiomatic loop:

``````  def to_i
first_element = @numbers.shift
first_number = ascii_value(first_element)

previous_number = first_number
until complete?
next_element = @numbers.shift
next_number = ascii_value(next_element)
previous_number = previous_number * 10 + next_number
end
previous_number
end

private

def ascii_value(n)
n.ord - '0'.ord
end

def complete?
!(@numbers.size > 0)
end
``````

# Exercises

1) Write the tests:

``````it "should convert '100' to 100"
it "should convert '1000' to 1000"
``````

Does it pass without making any modifications to the solution? Do you need these two tests?

2) Solve the problem without using Ruby's builtin bytes method.

# References

1. Table of ASCII Characters
2. How to Solve it Using Computers by R. G. Dromey

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