FizzBuzz/AWK
regular if / else
This is the "traditional" approach:
Loop, and modulo-check to see what to print.
Minor tweak: printf with no newline, and linebreaks only after each "FizzBuzz",
to get a more compact output.
<lang AWK># usage: awk -v n=38 -f FizzBuzz.awk
BEGIN {
if(!n) n=100 print "# FizzBuzz:"
for (ii=1; ii<=n; ii++) if (ii % 15 == 0) {print "FizzBuzz"} else if (ii % 3 == 0) {printf "Fizz "} else if (ii % 5 == 0) {printf "Buzz "} else {printf "%3d ", ii}
print "\n# Done."
}</lang>
- Output:
# FizzBuzz: 1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz Fizz 22 23 Fizz Buzz 26 Fizz 28 29 FizzBuzz 31 32 Fizz 34 Buzz Fizz 37 38 Fizz Buzz 41 Fizz 43 44 FizzBuzz 46 47 Fizz 49 Buzz Fizz 52 53 Fizz Buzz 56 Fizz 58 59 FizzBuzz 61 62 Fizz 64 Buzz Fizz 67 68 Fizz Buzz 71 Fizz 73 74 FizzBuzz 76 77 Fizz 79 Buzz Fizz 82 83 Fizz Buzz 86 Fizz 88 89 FizzBuzz 91 92 Fizz 94 Buzz Fizz 97 98 Fizz Buzz # Done.
When the output is presented like that, it is easy to see a pattern.
bash with echo
Using echo from the shell to generate the numbers as input.
Advantage: we need no loop inside the script.
Disadvantage: this needs a shell where echo can do this. <lang AWK>echo {1..100} | awk ' BEGIN {RS=" "} $1 % 15 == 0 {print "FizzBuzz"; next} $1 % 5 == 0 {printf "Buzz "; next} $1 % 3 == 0 {printf "Fizz "; next} {printf "%3d ",$1} '</lang>
One-liner with seq
Like version 2, using bash with seq to generate the numbers as input.
Disadvantage: needs external command seq, i.e. this only works on unix.
(Also, hard to read)
<lang AWK>seq 100 | awk '$0=NR%15?NR%5?NR%3?$0:"Fizz":"Buzz":"FizzBuzz"'</lang>
No divisions, using counters
Division is one of the more expensive operations, so it is nice if we can avoid it.
All processing is done inside awk, using no division & no modulo.
Instead, a simple counter for each of the output-variants is used:
<lang AWK># usage: awk -v n=38 -f fizzbuzzNoDiv.awk
- FizzBuzz using no division & no modulo-operations:
BEGIN {
if(!n) n=100 print "# FizzBuzz:" while (c1<n) {
c1++; c3++; c5++; cF++; x=sprintf("%3d ",c1) if(c3>= 3) { c3=0; x="Fizz " }
if(c5>= 5) { c5=0; x="Buzz " }
if(cF>=15) { cF=0; x="FizzBuzz\n" } printf(x)
} print "\n# Done."
}</lang> Same output as version 1.
No divisions, using pattern-string
Another solution that works without division / modulo.
This is inspired by the versions "Without Modulus" of Nimrod and Python,
using a precomputed (observed:) pattern to decide how to print each number.
But here, the pattern is represented as chars in a string,
instead of bits in an integer.
<lang AWK># usage: awk -v n=42 -f fizzbuzzRepeatPattern.awk
function prt(x,v) {
if(v==0) {printf("%3d ",x); return} # print number printf fb[v] # else: print text
} BEGIN {
if(!n) n=100 print "# FizzBuzz:"
pattern="003053003503006" # 0: print number, 3: print Fizz, etc. split("1,2, Fizz,4, Buzz, FizzBuzz\n,", fb, ",")
while (i<n) {
i++; sel++; prt(i, substr(pattern,sel,1) ); # select variant to use from the pattern
if( sel>=length(pattern) ) sel=0 } print "\n# Done."
}</lang> Same output as version 1.
Custom FizzBuzz
Example program generated from "General FizzBuzz", for factors 2, 3, 5 using the words A, B, C. <lang AWK># usage: awk -f fizzbuzzCustom.awk numbers.txt
BEGIN {print "# CustomFizzBuzz:"}
$1 % 2 == 0 {x = x "A"} $1 % 3 == 0 {x = x "B"} $1 % 5 == 0 {x = x "C"}
x=="" {print $0; next}
{print "\t" x; x=""}
END {print "# Done."} </lang>
- Output:
# CustomFizzBuzz: 1 A B A C AB 7 A B AC 11 AB 13 A BC A 17 AB 19 AC B A 23 AB C A B A 29 ABC 31 # Done.