When using regex in C, \d does not work but [0-9] does

Question

I do not understand why the regex pattern containing the \\d character class does not work but [0-9] does. Character classes, such as \\s

Answer 1

According to the POSIX regular expression spec:

An ordinary character is any character in the supported character set, except for the ERE special characters listed in ERE Special Characters. The interpretation of an ordinary character preceded by a backslash ( '\' ) is undefined.

So the only characters that can legally follow a \ are:

\^    \.    \[    \$    \(    \)    \|
\*    \+    \?    \{    \\

all of which match the escaped character literally. Trying to use any of of the other PCRE extensions may not work.

Answer 2

Trying either pattern in a strictly POSIX environment will likely end up having no matches; if you want to make the pattern truly POSIX compatible use all bracket expressions:

const char *rstr = "^[[:digit:]]+[[:space:]]+[[:alpha:]]+[[:space:]]+[[:digit:]]+[[:space:]]+[[:alpha:]]+$";

↳ POSIX Character_classes

Answer 3

\d is a perl and vim character class.

Use instead:

 const char *rstr = "^[[:digit:]]+\\s+\\w+\\s+[[:digit:]]+\\s+\\w+$"; 

Answer 4

The regex flavor you're using is GNU ERE, which is similar to POSIX ERE, but with a few extra features. Among these are support for the character class shorthands \s, \S, \w and \W, but not \d and \D. You can find more info here.