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#! /usr/bin/env gforth
\ Challenge 001
\
\ Challenge #2
\ Write a one-liner to solve the FizzBuzz problem and print the numbers 1 through 20. However, any number divisible by 3 should be replaced by the word �fizz� and any divisible by 5 by the word �buzz�. Those numbers that are both divisible by 3 and 5 become �fizzbuzz�.
: fizzbuzz ( n -- )
1+ 1 ?DO
I 15 MOD 0= IF ." fizzbuzz" ELSE
I 3 MOD 0= IF ." fizz" ELSE
I 5 MOD 0= IF ." buzz" ELSE I .
THEN THEN THEN
CR
LOOP
;
20 fizzbuzz
BYE
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