One thing you are missing is that (\d?) doesn't fail and backtrack as you describe; it succeeds in matching the "1" in both cases and then applies the rest of the regex (\.?)(\d?|\d+)$ to what comes after the "1". Only if that rest of the regex fails will it try first having \d? match 0 digits and applying the rest of the regex to the whole string and then the \d+ alternative, matching first as many digits as possible, then successively fewer until the end of the regex matches.

But for "12345", it does almost no backtracking; \d? matches the "1", \.? doesn't match once but succeeds at matching 0 times, the second \d? matches the "2", but the $ doesn't match so \d? tries matching 0 times, $ still doesn't match, so the second \d+ is tried, matching "2345", and then $ matches the end of string.

(\d?|\d+) is a very strange construct; it says to try matching in this order: 1 digit, 0 digits, N digits, N-1 digits, N-2 digits, ..., 2 digits. I can't believe that's really what you want. Do you want something as simple as: <c>/^(-?)(\d+)(\.?)(\d*)$/<c>

In reply to Re^3: combined into a single regex by ysth
in thread combined into a single regex by arcnon

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