Is it possible to increment numbers using regex substitution?

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This question’s topic amused me for one particular implementation I did earlier. My solution happens to be two substitutions so I’ll post it.

My implementation environment is solaris, full example:

echo "0 1 2 3 7 8 9 10 19 99 109 199 909 999 1099 1909" |
perl -pe 's/\b([0-9]+)\b/0$1~01234567890/g' |
perl -pe 's/\b0(?!9*~)|([0-9])(?=9*~[0-9]*?\1([0-9]))|~[0-9]*/$2/g'

1 2 3 4 8 9 10 11 20 100 110 200 910 1000 1100 1910

Pulling it apart for explanation:

s/\b([0-9]+)\b/0$1~01234567890/g

For each number (#) replace it with 0#~01234567890. The first 0 is in case rounding 9 to 10 is needed. The 01234567890 block is for incrementing. The example text for “9 10” is:

09~01234567890 010~01234567890

The individual pieces of the next regex can be described seperately, they are joined via pipes to reduce substitution count:

s/\b0(?!9*~)/$2/g

Select the “0” digit in front of all numbers that do not need rounding and discard it.

s/([0-9])(?=9*~[0-9]*?\1([0-9]))/$2/g

(?=) is positive lookahead, \1 is match group #1. So this means match all digits that are followed by 9s until the ‘~’ mark then go to the lookup table and find the digit following this number. Replace with the next digit in the lookup table. Thus “09~” becomes “19~” then “10~” as the regex engine parses the number.

s/~[0-9]*/$2/g

This regex deletes the ~ lookup table.

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