This is a question in my paper test today, the function signature is
int is_match(char* pattern,char* string)
The pattern is limited to only ASCII chars and the quantification *
and ?
, so it is relatively simple. is_match
should return 1 if matched, otherwise 0.
How do I do this?
Brian Kernighan provided a short article on A Regular Expression Matcher that Rob Pike wrote as a demonstration program for a book they were working on. The article is a very nice read explaining a bit about the code and regular expressions in general.
I have played with this code, making a few changes to experiment with some extensions such as to also return where in the string the pattern matches so that the substring matching the pattern can be copied from the original text.
From the article:
I suggested to Rob that we needed to find the smallest regular expression package that would illustrate the basic ideas while still recognizing a useful and non-trivial class of patterns. Ideally, the code would fit on a single page.
Rob disappeared into his office, and at least as I remember it now, appeared again in no more than an hour or two with the 30 lines of C code that subsequently appeared in Chapter 9 of TPOP. That code implements a regular expression matcher that handles these constructs:
c matches any literal character c . matches any single character ^ matches the beginning of the input string $ matches the end of the input string * matches zero or more occurrences of the previous character
This is quite a useful class; in my own experience of using regular expressions on a day-to-day basis, it easily accounts for 95 percent of all instances. In many situations, solving the right problem is a big step on the road to a beautiful program. Rob deserves great credit for choosing so wisely, from among a wide set of options, a very small yet important, well-defined and extensible set of features.
Rob's implementation itself is a superb example of beautiful code: compact, elegant, efficient, and useful. It's one of the best examples of recursion that I have ever seen, and it shows the power of C pointers. Although at the time we were most interested in conveying the important role of a good notation in making a program easier to use and perhaps easier to write as well, the regular expression code has also been an excellent way to illustrate algorithms, data structures, testing, performance enhancement, and other important topics.
The actual C source code from the article is very very nice.
/* match: search for regexp anywhere in text */
int match(char *regexp, char *text)
{
if (regexp[0] == '^')
return matchhere(regexp+1, text);
do { /* must look even if string is empty */
if (matchhere(regexp, text))
return 1;
} while (*text++ != '\0');
return 0;
}
/* matchhere: search for regexp at beginning of text */
int matchhere(char *regexp, char *text)
{
if (regexp[0] == '\0')
return 1;
if (regexp[1] == '*')
return matchstar(regexp[0], regexp+2, text);
if (regexp[0] == '$' && regexp[1] == '\0')
return *text == '\0';
if (*text!='\0' && (regexp[0]=='.' || regexp[0]==*text))
return matchhere(regexp+1, text+1);
return 0;
}
/* matchstar: search for c*regexp at beginning of text */
int matchstar(int c, char *regexp, char *text)
{
do { /* a * matches zero or more instances */
if (matchhere(regexp, text))
return 1;
} while (*text != '\0' && (*text++ == c || c == '.'));
return 0;
}
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