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A fundamental feature of regular expression matching involves the notion called backtracking. which is used (when needed) by all regular expression quantifiers, namely *, *?, +, +?, {n,m}, and {n,m}?.

For a regular expression to match, the entire regular expression must match, not just part of it. So if the beginning of a pattern containing a quantifier succeeds in a way that causes later parts in the pattern to fail, the matching engine backs up and recalculates the beginning part-that's why it's called backtracking.

Here is an example of backtracking: Let's say you want to find the word following "foo" in the string "Food is on the foo table.":

    $_ = "Food is on the foo table.";
    if ( /\b(foo)\s+(\w+)/i ) {
        print "$2 follows $1.\n";

When the match runs, the first part of the regular expression (\b(foo)) finds a possible match right at the beginning of the string, and loads up $1 with "Foo". However, as soon as the matching engine sees that there's no whitespace following the "Foo" that it had saved in $1, it realizes its mistake and starts over again one character after where it had the tentative match. This time it goes all the way until the next occurrence of "foo". The complete regular expression matches this time, and you get the expected output of "table follows foo."

Sometimes minimal matching can help a lot. Imagine you'd like to match everything between "foo" and "bar". Initially, you write something like this:

    $_ =  "The food is under the bar in the barn.";
    if ( /foo(.*)bar/ ) {
        print "got <$1>\n";

Which perhaps unexpectedly yields:

  got <d is under the bar in the >

That's because .* was greedy, so you get everything between the first "foo" and the last "bar". In this case, it's more effective to use minimal matching to make sure you get the text between a "foo" and the first "bar" thereafter.

    if ( /foo(.*?)bar/ ) { print "got <$1>\n" }
  got <d is under the >

Here's another example: let's say you'd like to match a number at the end of a string, and you also want to keep the preceding part the match. So you write this:

    $_ = "I have 2 numbers: 53147";
    if ( /(.*)(\d*)/ ) {                                # Wrong!
        print "Beginning is <$1>, number is <$2>.\n";

That won't work at all, because .* was greedy and gobbled up the whole string. As \d* can match on an empty string the complete regular expression matched successfully.

    Beginning is <I have 2 numbers: 53147>, number is <>.

Here are some variants, most of which don't work:

    $_ = "I have 2 numbers: 53147";
    @pats = qw{

    for $pat (@pats) {
        printf "%-12s ", $pat;
        if ( /$pat/ ) {
            print "<$1> <$2>\n";
        } else {
            print "FAIL\n";

That will print out:

    (.*)(\d*)    <I have 2 numbers: 53147> <>
    (.*)(\d+)    <I have 2 numbers: 5314> <7>
    (.*?)(\d*)   <> <>
    (.*?)(\d+)   <I have > <2>
    (.*)(\d+)$   <I have 2 numbers: 5314> <7>
    (.*?)(\d+)$  <I have 2 numbers: > <53147>
    (.*)\b(\d+)$ <I have 2 numbers: > <53147>
    (.*\D)(\d+)$ <I have 2 numbers: > <53147>

As you see, this can be a bit tricky. It's important to realize that a regular expression is merely a set of assertions that gives a definition of success. There may be 0, 1, or several different ways that the definition might succeed against a particular string. And if there are multiple ways it might succeed, you need to understand backtracking to know which variety of success you will achieve.

When using lookahead assertions and negations, this can all get even tricker. Imagine you'd like to find a sequence of non-digits not followed by "123". You might try to write that as

        $_ = "ABC123";
        if ( /^\D*(?!123)/ ) {                          # Wrong!
            print "Yup, no 123 in $_\n";

But that isn't going to match; at least, not the way you're hoping. It claims that there is no 123 in the string. Here's a clearer picture of why it that pattern matches, contrary to popular expectations:

    $x = 'ABC123' ;
    $y = 'ABC445' ;

    print "1: got $1\n" if $x =~ /^(ABC)(?!123)/ ;
    print "2: got $1\n" if $y =~ /^(ABC)(?!123)/ ;

    print "3: got $1\n" if $x =~ /^(\D*)(?!123)/ ;
    print "4: got $1\n" if $y =~ /^(\D*)(?!123)/ ;

This prints

    2: got ABC
    3: got AB
    4: got ABC

You might have expected test 3 to fail because it seems to a more general purpose version of test 1. The important difference between them is that test 3 contains a quantifier (\D*) and so can use backtracking, whereas test 1 will not. What's happening is that you've asked "Is it true that at the start of $x, following 0 or more non-digits, you have something that's not 123?" If the pattern matcher had let \D* expand to "ABC", this would have caused the whole pattern to fail. The search engine will initially match \D* with "ABC". Then it will try to match (verb|!123| with "123" which, of course, fails. But because a quantifier (\D*) has been used in the regular expression, the search engine can backtrack and retry the match differently in the hope of matching the complete regular expression.

Well now, the pattern really, really wants to succeed, so it uses the standard regular expression back-off-and-retry and lets \D* expand to just "AB" this time. Now there's indeed something following "AB" that is not "123". It's in fact "C123", which suffices.

We can deal with this by using both an assertion and a negation. We'll say that the first part in $1 must be followed by a digit, and in fact, it must also be followed by something that's not "123". Remember that the lookaheads are zero-width expressions-they only look, but don't consume any of the string in their match. So rewriting this way produces what you'd expect; that is, case 5 will fail, but case 6 succeeds:

    print "5: got $1\n" if $x =~ /^(\D*)(?=\d)(?!123)/ ;
    print "6: got $1\n" if $y =~ /^(\D*)(?=\d)(?!123)/ ;

    6: got ABC

In other words, the two zero-width assertions next to each other work like they're ANDed together, just as you'd use any builtin assertions: /^$/ matches only if you're at the beginning of the line AND the end of the line simultaneously. The deeper underlying truth is that juxtaposition in regular expressions always means AND, except when you write an explicit OR using the vertical bar. /ab/ means match "a" AND (then) match "b", although the attempted matches are made at different positions because "a" is not a zero-width assertion, but a one-width assertion.

One warning: particularly complicated regular expressions can take exponential time to solve due to the immense number of possible ways they can use backtracking to try match. For example this will take a very long time to run


And if you used *'s instead of limiting it to 0 through 5 matches, then it would take literally forever-or until you ran out of stack space.

There are some other pattern matching constructs see recommended Perl books or online Perl references for more details and plenty of examples.

next up previous contents
Next: Setting the Target Operator Up: Regular Expressions Previous: Special pattern matching character