Visual Studio replace magic number integers with doubles

Question

Can a regular expression be used to find all magic number integers in Visual Studio and convert them to doubles?

The regular expression to find them is beyond my skill, here is what I came up with so far:

(?!=[\s\(])(?<!\.)\d+(?=[\s\)])

This will find all the integers, but erroneously matches

15.527 = "1"5.5"2""7"

Even with a perfect match, is there a way to replace with same number? For example 7 would be replaced by 7.0 and 16 would be replaced by 16.0

Can you explain this? Given `P3-4-5` it matches 4 and 5, and I'd like to ignore them but don't understand the expression. — Adam, Feb 27 '17 at 17:20
In theory you have to match all the pragma's, define's, include's, or any custom MS directives. Then you also have to match all quotes and comments. Once you are matching that, you can tack on a sub expression to match numbers. And the numbers can be separated by whitespace. — , Feb 27 '17 at 17:22
See https://regex101.com/r/9J6HGv/1, there is an explanation on the right. — Wiktor Stribiżew, Feb 27 '17 at 17:25
Why do you need to do this? What problem are you trying to solve in which this is the solution, or on the path to the solution? It sounds like an X/Y problem to be honest. — Lasse V. Karlsen, Feb 27 '17 at 21:15

score 1 · Answer 1 · answered Feb 27 '17 at 17:41

1

Using the regex provided by @Wiktor Stribiżew

answered Feb 27 '17 at 17:41

Adam

Why did you post this if you said it does not work for you? I could do it myself if you said otherwise. – Wiktor Stribiżew Feb 27 '17 at 18:40
@WiktorStribiżew it did work for me, it just matched a few extra things that I manually reverted. – Adam Feb 27 '17 at 20:45

score 1 · Accepted Answer · edited May 23 '17 at 12:00

It seems the pattern I suggested works in almost 99% of cases, so, let me explain the (?<![0-9]\.)\b[0-9]+\b(?!\.) pattern:

(?<![0-9]\.) - a negative lookbehind failing the match if there is a digit and a . symbols right before the digit matched with [0-9]+ pattern
\b - a leading word boundary (i.e. the previous char cannot be a letter/digit/underscore)
[0-9]+ - 1 or more digits
\b - a trailing word boundary (i.e. the next char after the last matched digit cannot be a letter/digit/underscore)
(?!\.) - a negative lookahead that fails the match if the next char after the last digit matched by the [0-9]+ subpattern is followed with ..

Why [0-9]+ instead of \d? See this SO thread, \d is Unicode aware and matches more than just 0..9 digits.

I am not sure the `(?<![0-9]\.)\b[0-9]+\b(?![-\s.]|$)` regex will work in 100% cases either, it is just an "extension" of the one above: the lookahead fails the matches even when the digits are followed with `-` or whitespace (`[-\s]`), or if they are at the end of string (`$`). — Wiktor Stribiżew, Feb 27 '17 at 21:21

2 Answers2