How to test that a string follows a given pattern in C?

Question

I would like to check whether a string follows a certain pattern. I tried sscanf, but I do not get the desired result.

The pattern is simple: it consists of:

the string "while" followed by
one or more spaces, followed by
a string made of alpha characters or the underscore character, followed by
zero or more spaces, followed by
a colon (':'), followed by
the newline character ('\n')

Pattern examples:

while condition_a:
while test_b :

I tried the following, but it would not check for the column:

sscanf(string, "while %[a-z,_]s %[:]c", test, column);

Would you have any suggestion?

I think you need a C regex library. Check out this post: http://stackoverflow.com/q/1085083/10077 — Fred Larson, Jan 03 '14 at 20:34

score 6 · Accepted Answer · 2014-01-03T21:16:11.590

Seems pretty straightforward to implement. You don't need either the unintuitive and quirky scanf(), nor non-portable (and, frankly, horrible) regular expressions:

int isValid(const char *s)
{
    // the string "while" followed by
    if (memcmp(s, "while", 5))
        return 0;

    s += 5;

    // one or more spaces, followed by
    if (!isspace(*s))
        return 0;

    while (isspace(*++s))
        ;

    // a string made of alpha characters or the underscore character,
    // (I assumed zero or more)
    while (isalpha(*s) || *s == '_')
        s++;

    // followed by zero or more spaces
    while (isspace(*s))
        s++;

    // followed by a column (':'),
    if (*s++ != ':')
        return 0;

    // followed by the newline character ('\n')
    if (*s++ != '\n')
        return 0;

    // here should be the end
    return !*s;
}

Thanks, I used your method. – lcazarre Jan 04 '14 at 11:37 — lcazarre, Jan 04 '14 at 11:37

Joseph Quinsey · Answer 2 · 2014-01-24T21:28:36.197

1

This test for your pattern seems to work:

   int n = 0;
   Bool ok = sscanf(string, "while%*[ ]%*[A-Za-z_] :%*1[\n]%n", &n) == 0 && 
      n && !string[n];

It is nice and short, but there are (at least) two flaws:

It is ugly
It allows arbitrary whitespace before the colon, not just spaces (e.g. tabs, newlines)

The only way to handle zero or more spaces in sscanf is to use it twice, once for one or more, and then again for zero. For example, this code:

   char tail[4] = "";
   Bool ok = (sscanf(string, "while%*[ ]%*[A-Za-z_]%*[ ]%3c", tail) == 1 || 
              sscanf(string, "while%*[ ]%*[A-Za-z_]%3c",      tail) == 1) && 
              !strcmp(tail, ":\n");

edited Jan 24 '14 at 21:28

answered Jan 03 '14 at 20:46

Joseph Quinsey

9,553
10
54
77

You should point out that the 'scan set' (`%[]`) is a conversion specification in its own right; you don't need to specify a `s` or a `c` after it. Your code handles that. You also insist on two spaces after `while` because `sizeof("while")` is 6 (the null at the end is counted). – Jonathan Leffler Jan 03 '14 at 21:16

Bill Lynch · Answer 3 · 2014-01-03T21:20:30.147

A regular expression seems like a reasonable tool here:

#include <assert.h>
#include <regex.h>
#include <stdio.h>
#include <stdlib.h>

int main() {
    const char *expression = "^while +([a-zA-Z_]+) *:\n$";
    const char *input = NULL;
    regex_t regex;
    int rc;

    size_t nmatch = 2;
    regmatch_t pmatch[2];

    rc = regcomp(&regex, expression, REG_EXTENDED);
    assert(rc == 0);

    input = "while condition_a:\n";
    rc = regexec(&regex, input, nmatch, pmatch, 0);
    if(rc == 0) {
        printf("Match: %.*s\n", (int)(pmatch[1].rm_eo - pmatch[1].rm_so), input + pmatch[1].rm_so);
    } else if (rc == REG_NOMATCH) {
        printf("No match\n");
    } else {
        char msgbuf[64];
        regerror(rc, &regex, msgbuf, sizeof(msgbuf));
        printf("Regex match failed: %s\n", msgbuf);
    }

    input = "while test_b :\n";
    rc = regexec(&regex, input, nmatch, pmatch, 0);
    if(rc == 0) {
        printf("Match: %.*s\n", (int)(pmatch[1].rm_eo - pmatch[1].rm_so), input + pmatch[1].rm_so);
    } else if (rc == REG_NOMATCH) {
        printf("No match\n");
    } else {
        char msgbuf[64];
        regerror(rc, &regex, msgbuf, sizeof(msgbuf));
        printf("Regex match failed: %s\n", msgbuf);
    }

    regfree(&regex);
}

This will output:

Match: condition_a
Match: test_b

How to test that a string follows a given pattern in C?

3 Answers3