stop condition for emulating "grep -oE" with awk - regex

Question

I'm trying to emulate GNU grep -Eo with a standard awk call.
What the man says about the -o option is:
-o --only-matching
     Print only the matched (non-empty) parts of matching lines, with each such part on a separate output line.
For now I have this code:
#!/bin/sh
regextract() {
[ "$#" -ge 2 ] || return 1
__regextract_ere=$1
shift
awk -v FS='^$' -v ERE="$__regextract_ere" '
{
while ( match($0,ERE) && RLENGTH > 0 ) {
print substr($0,RSTART,RLENGTH)
$0 = substr($0,RSTART+1)
}
}
' "$#"
}
My question is: In the case that the matching part is 0-length, do I need to continue trying to match the rest of the line or should I move to the next line (like I already do)? I can't find a sample of input+regex that would need the former but I feel like it might exist. Any idea?

Answer 1

Here's a POSIX awk version, which works with a* (or any POSIX awk regex):
echo abcaaaca |
awk -v regex='a*' '
{
while (match($0, regex)) {
if (RLENGTH) print substr($0, RSTART, RLENGTH)
$0 = substr($0, RSTART + (RLENGTH > 0 ? RLENGTH : 1))
if ($0 == "") break
}
}'
Prints:
a
aaa
a
POSIX awk and grep -E use POSIX extended regular expressions, except that awk allows C escapes (like \t) but grep -E does not. If you wanted strict compatibility you'd have to deal with that.

Answer 2

If you can consider a gnu-awk solution then using RS and RT may give identical behavior of grep -Eo.
# input data
cat file
FOO:TEST3:11
BAR:TEST2:39
BAZ:TEST0:20
Using grep -Eo:
grep -Eo '[[:alnum:]]+' file
FOO
TEST3
11
BAR
TEST2
39
BAZ
TEST0
20
Using gnu-awk with RS and RT using same regex:
awk -v RS='[[:alnum:]]+' 'RT != "" {print RT}' file
FOO
TEST3
11
BAR
TEST2
39
BAZ
TEST0
20
More examples:
grep -Eo '\<[[:digit:]]+' file
11
39
20
awk -v RS='\\<[[:digit:]]+' 'RT != "" {print RT}' file
11
39
20

Answer 3

Thanks to the various comments and answers I think that I have a working, robust, and (maybe) efficient code now:
tested on AIX/Solaris/FreeBSD/macOS/Linux
#!/bin/sh
regextract() {
[ "$#" -ge 1 ] || return 1
[ "$#" -eq 1 ] && set -- "$1" -
awk -v FS='^$' '
BEGIN {
ere = ARGV[1]
delete ARGV[1]
}
{
tail = $0
while ( tail != "" && match(tail,ere) ) {
if (RLENGTH) {
print substr(tail,RSTART,RLENGTH)
tail = substr(tail,RSTART+RLENGTH)
} else
tail = substr(tail,RSTART+1)
}
}
' "$#"
}
regextract "$#"
notes:
I pass the ERE string along the file arguments so that awk doesn't pre-process it (thanks #anubhava for pointing that out); C-style escape sequences will still be translated by the regex engine of awk though (thanks #dan for pointing that out).
Because assigning $0 does reset the values of all fields,
I chose FS = '^$' for limiting the overhead
Copying $0 in a separate variable nullifies the overhead induced by assigning $0 in the while loop (thanks #EdMorton for pointing that out).
a few examples:
# Multiple matches in a single line:
echo XfooXXbarXXX | regextract 'X*'
X
XX
XXX
# Passing the regex string to awk as a parameter versus a file argument:
echo '[a]' | regextract_as_awk_param '\[a]'
a
echo '[a]' | regextract '\[a]'
[a]
# The regex engine of awk translates C-style escape sequences:
printf '%s\n' '\t' | regextract '\t'
printf '%s\n' '\t' | regextract '\\t'
\t

Answer 4

Your code will malfunction for match which might have zero or more characters, consider following simple example, let file.txt content be
1A2A3
then
grep -Eo A* file.txt
gives output
A
A
your while's condition is match($0,ERE) && RLENGTH > 0, in this case former part gives true, but latter gives false as match found is zero-length before first character (RSTART was set to 1), thus body of while will be done zero times.