How does {m}{n} ("exactly n times" twice) work?

Question

So, some way or another (playing around), I found myself with a regex like \d{1}{2}.

Logically, to me, it should mean:

(A digit exactly once) exactly twice, i.e. a digit exactly twice.

But it, in fact, appears to just mean "a digit exactly once" (thus ignoring the {2}).

String regex = "^\\d{1}{2}$"; // ^$ to make those not familiar with 'matches' happy System.out.println("1".matches(regex)); // true System.out.println("12".matches(regex)); // false 

Similar results can be seen using {n}{m,n} or similar.

Why does this happen? Is it explicitly stated in regex / Java documentation somewhere or is it just a decision Java developers made on-the-fly or is it maybe a bug?

Or is it in fact not ignored and it actually means something else entirely?

Not that it matters much, but it's not across-the-board regex behaviour, Rubular does what I expect.

Note - the title is mainly for searchability for users who want to know how it works (not why).

Answer 1

IEEE-Standard 1003.1 says:

The behavior of multiple adjacent duplication symbols ( '*' and intervals) produces undefined results.

So every implementation can do as it pleases, just don't rely on anything specific...

Answer 2

When I input your regex in RegexBuddy using the Java regex syntax, it displays following message

Quantifiers must be preceded by a token that can be repeated «{2}»

Changing the regex to explicitly use a grouping ^(\d{1}){2} solves that error and works as you expect.

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I assume that the java regex engine simply neglects the error/expression and works with what has been compiled so far.

Edit

The reference to the IEEE-Standard in @piet.t's answer seems to support that assumption.

Edit 2 (kudos to @fncomp)

For completeness, one would typically use (?:)to avoid capturing the group. The complete regex then becomes ^(?:\d{1}){2}