#ElixirGolf FizzBuzz: https://twitter.com/elixirgolf/status/665139870818574336

1-fizzbuzz.exs

# Shortest (91 chars):
for n<-1..?d,r=&(rem(n,&1)==0&&IO.write&2)do;f=r.(3,"Fizz");b=r.(5,"Buzz");f||b||r.(1,n)end

# If newlines are required (99 chars):
for n<-1..?d,r=&(rem(n,&1)==0&&IO.write&2),do: (f=r.(3,"Fizz");b=r.(5,"Buzz");f||b||r.(1,n);:io.nl)

# Other solutions:
for n<-1..?d,r=&(rem(n,&1)==0&&IO.puts&2),do: r.(15,"FizzBuzz")||r.(3,"Fizz")||r.(5,"Buzz")||r.(1,n)
for n<-1..?d,do: Enum.find [{15,"FizzBuzz"},{3,"Fizz"},{5,"Buzz"},{1,n}],fn({d,t})->rem(n,d)==0&&IO.puts(t)end

2-notes.md

• ?d is 100 (ASCII code for d)
• for …,this,do abuses a for-comprehension filter. this runs on each iteration and would skip if returning a falsy value.
• Since IO.write returns a truthy value, we can chain f||b||r.(1,n) to only output the bare number if Fizz or Buzz did not output.

Great... thanks for the descriptions really helpful - also like the conditions of 1==1&&IO.write "match"

Also see @zyro's: https://gist.github.com/zyro/e60e75ef3a3c3f08e5ef