AWK function w/ a variable number of arguments - awk

Question

How to define an AWK function w/ a variable number of arguments? I can emulate this w/ command line arguments:
awk 'BEGIN {for (i in ARGV) printf ARGV[i]" "}' 1 2 3
but BEGIN {for (i in ARGV) printf ARGV[i]" "} isn't a function (in AWK).
Currently I'm using MAWK (but can probably switch to GAWK if it would help)
NOTE: I cannot reveal the task (it's an exercise which I'm supposed to solve by myself)

Answer 1

There are no variadic functions in awk because they're not needed since you can just populate an array and pass that into a function:
$ cat tst.awk
BEGIN {
split("foo 17 bar",a)
foo(a)
}
function foo(arr, i,n) {
n = length(arr) # or loop incrementing n if length(arr) unsupported
for (i=1; i<=n; i++) {
printf "%s%s", arr[i], (i<n ? OFS : ORS)
}
}
$ awk -f tst.awk
foo 17 bar
or just define the function with a bunch of dummy argument names as #triplee mentions.

Answer 2

As per https://rosettacode.org/wiki/Variadic_function you can define a function with more arguments than you pass in; the ones you omit will come out as empty.
It's not clear from your example what you are actually trying to accomplish, but this is the answer to your actual question.
$ awk 'function handlemany(first, second, third, fourth, fifth, sixth, seventh, eighth, ninth, tenth, eleventh, twelfth) {
> print first, second, third, fourth, fifth, sixt, seventh, eighth, ninth, tenth, eleventh, twelfth
> }
> BEGIN { handlemany("one", "two", "three") }'
one two three
This is less than ideal, of course, but there is no support for proper variadic functions / varargs in the Awk language.