Instructions
Use the IO.gets function to gather an integer from user input and return
a string where EACH digit is converted to an "e" or an "o" string character.
This string character is determined by whether the integer digit at that index
is odd or even.
For example:
iex> --enter your code here--
13387
"oooeo"Solutions
Winner
The winning solution was submitted by @henrik:
char count: 66
iex(1)> "#{for x<-(to_char_list IO.gets''),x>10,do: Enum.at'eo',rem(x,2)}"
12231
"oeeoo"Starting in the middle it IO.gets the user’s input, and converts this binary
string into a character list (a list of integer values). Note that the last
character will be a \n which is ASCII value 10.
It uses a comprehension (for x<-...) to take each integer from our list and
apply the x>10 filter. As all the string values are greater than 10 they
pass through, where as the \n escapes. The values are then passed to the body
of our comprehension do: Enum.at....
In the body, the integer value bound to x is divided by 2 determining whether it is even or
odd. This remainder value is then used by Enum.at to return the
corresponding "e" or "o" character.
Finally the code is wrapped in an interpolated string "#{}" to convert the
character list into a binary string.
Honourable mentions
Submitted by @henrik:
"#{for x<-'13387',do: Enum.at'eo',rem(x,2)}""#{Enum.map'13387',&%{0=>?e,1=>?o}[rem(&1,2)]}""#{for x<-'13387',do: Enum.at'eoeoeoeoeo',x-48}"Submitted by @meadsteve:
import String;for s<-:stdio|>IO.gets|>codepoints,s !="\n",x=s|>to_integer|>rem(2),do: Enum.at'eo',xResources
- Elixir docs
Enum.at: http://elixir-lang.org/docs/stable/elixir/Enum.html#at/3 - Comprehension docs
for: http://elixir-lang.org/getting-started/comprehensions.html - More comprehension docs: http://elixir-lang.org/docs/stable/elixir/Kernel.SpecialForms.html#for/1
- String interpolation: http://elixir-lang.org/getting-started/basic-types.html#strings