Regular Expression replacing pattern with pattern

Question

I am new to regular expressions. I was googling and found some solutions and then I come up with my own solution as following

#include <string.h>
#include <regex.h>
#include <iostream>

int rreplace (char *buf, int size, regex_t *re, char *rp){
    char *pos;
    int sub, so, n;
    regmatch_t pmatch [10];
    if (regexec (re, buf, 10, pmatch, 0))
        return 0;
    for (pos = rp; *pos; pos++)
        if (*pos == '\\' && *(pos + 1) > '0' && *(pos + 1) <= '9'){
            so = pmatch [*(pos + 1) - 48].rm_so;
            n = pmatch [*(pos + 1) - 48].rm_eo - so;
            if (so < 0 || strlen (rp) + n - 1 > size)
                return 1;
            memmove (pos + n, pos + 2, strlen (pos) - 1);
            memmove (pos, buf + so, n);
            pos = pos + n - 2;
        }

    sub = pmatch [1].rm_so; /* no repeated replace when sub >= 0 */
    for (pos = buf; !regexec (re, pos, 1, pmatch, 0); ){
        n = pmatch [0].rm_eo - pmatch [0].rm_so;
        pos += pmatch [0].rm_so;
        if (strlen (buf) - n + strlen (rp) + 1 > size)
            return 1;
        memmove (pos + strlen (rp), pos + n, strlen (pos) - n + 1);
        memmove (pos, rp, strlen (rp));
        pos += strlen (rp);
        if (sub >= 0)
            break;
    }
    return 0;
}

int main (int argc, char **argv){
    //buf [FILENAME_MAX],
    char rp [FILENAME_MAX];
    regex_t re;
    string toBeReplaced = "-";
     string replacedWith = "/";
    regcomp (&re, toBeReplaced.c_str(), REG_ICASE);

    string buf;
    cout << "Enter date separated with dash" << endl;
    cin >> buf;

    char * replacedWith_ = new char[replacedWith.size() + 1];
    std::copy(replacedWith.begin(), replacedWith.end(), replacedWith_);
    replacedWith_[replacedWith.size()] = '\0'; // don't forget the terminating 0


    char * buf_ = new char[buf.size() + 1];
    std::copy(buf.begin(), buf.end(), buf_);
    buf_[buf.size()] = '\0'; // don't forget the terminating 0


    rreplace (buf_, FILENAME_MAX, &re, strcpy (rp, replacedWith_));

    cout<<  buf_ << endl;
    regfree (&re);
    delete[] replacedWith_;
    return 0;
}

Well this code works fine if my string contains something like

22-04-2013

and it will change it to

22/04/2013. but I want it to be generic something like

\d\d-\d\d-\d\d\d\d

to be replaced with

\d\d/\d\d/\d\d\d\d

as I want it to be generic. Also I am working in linux g++. Most of the on-line solutions available are on different platforms. I also tried the following

string toBeReplaced = "\d[-]\d";
&
string replacedWith = "\d/\d";

but no luck. and I get \d/\d when I enter 3-4. I dont know why. Forgive me if I asked something stupid.

EDIT

My problem is match a pattern and replace it with a pattern. like digit followed by a hyphen should be replaced with digit followed by a slash.

Answer 1

Problem

You cannot replace a match with more regex, you will get the literal text \d/\d instead.

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Solution

To accomplish your goal, you need replace using a backreferenced capture group() like so:

(\d{2})-(\d{2})-(\d{4})

Your replacement string would be as follows:

$1/$2/$3

As you can tell, each capture group is numbered. There are three capture groups in the above regex.

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Further notes on Capture Groups

• A numbered capture group is specified whenever you wrap an expression in (regex)
• To keep your expression orderly, you can specify non-capturing groups like this (?:regex)
• For easier backreferencing, create a named capture group like this: (?<name>regex)
• To refer to a named capture group in a replacement string, use ${name}, instead of using $1

Please note in the above examples, regex should be replaced by your desired Regular Expression.

Some syntax for backrefencing may vary with different Regex Implementations, for example: \1 instead of $1

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Demonstration

Here's a demo for a visual representation of what I'm talking about:

Regex101 Example, Capture Groups on a Datetime String