c++ regex extract all substrings using regex_search()

Question

I am new to c++ regex. I have a string "{1,2,3}" and I want to extract the numbers 1 2 3. I thought I should use regex_search but it failed.

#include<iostream>
#include<regex>
#include<string>
using namespace std;
int main()
{
        string s1("{1,2,3}");
        string s2("{}");
        smatch sm;
        regex e(R"(\d+)");
        cout << s1 << endl;
        if (regex_search(s1,sm,e)){
                cout << "size: " << sm.size() << endl;
                for (int i = 0 ; i < sm.size(); ++i){
                        cout << "the " << i+1 << "th match" <<": "<< sm[i] <<  endl;
                }
        }
}

The result:

{1,2,3}
size: 1
the 1th match: 1

You probably want to use regex_match() with an appropriate expression describing this format. — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Sep 13, 2015 at 19:29

Galik · Accepted Answer · 2017-05-15 12:48:27Z

24

std::regex_search returns after only the first match found.

What std::smatch gives you is all the matched groups in the regular expression. Your regular expression only contains one group so std::smatch only has one item in it.

If you want to find all matches you need to use std::sregex_iterator.

int main()
{
    std::string s1("{1,2,3}");
    std::regex e(R"(\d+)");

    std::cout << s1 << std::endl;

    std::sregex_iterator iter(s1.begin(), s1.end(), e);
    std::sregex_iterator end;

    while(iter != end)
    {
        std::cout << "size: " << iter->size() << std::endl;

        for(unsigned i = 0; i < iter->size(); ++i)
        {
            std::cout << "the " << i + 1 << "th match" << ": " << (*iter)[i] << std::endl;
        }
        ++iter;
    }
}

Output:

{1,2,3}
size: 1
the 1th match: 1
size: 1
the 1th match: 2
size: 1
the 1th match: 3

The end iterator is default constructed by design so that it is equal to iter when iter has run out of matches. Notice at the bottom of the loop I do ++iter. That moves iter on to the next match. When there are no more matches, iter has the same value as the default constructed end.

Another example to show the submatching (capture groups):

int main()
{
    std::string s1("{1,2,3}{4,5,6}{7,8,9}");
    std::regex e(R"~((\d+),(\d+),(\d+))~");

    std::cout << s1 << std::endl;

    std::sregex_iterator iter(s1.begin(), s1.end(), e);
    std::sregex_iterator end;

    while(iter != end)
    {
        std::cout << "size: " << iter->size() << std::endl;

        std::cout << "expression match #" << 0 << ": " << (*iter)[0] << std::endl;
        for(unsigned i = 1; i < iter->size(); ++i)
        {
            std::cout << "capture submatch #" << i << ": " << (*iter)[i] << std::endl;
        }
        ++iter;
    }
}

Output:

{1,2,3}{4,5,6}{7,8,9}
size: 4
expression match #0: 1,2,3
capture submatch #1: 1
capture submatch #2: 2
capture submatch #3: 3
size: 4
expression match #0: 4,5,6
capture submatch #1: 4
capture submatch #2: 5
capture submatch #3: 6
size: 4
expression match #0: 7,8,9
capture submatch #1: 7
capture submatch #2: 8
capture submatch #3: 9

edited May 15, 2017 at 12:48

answered Sep 13, 2015 at 19:31

Galik

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2 Comments

daydayup Over a year ago

Thank you. Can I ask what "end" is for? Looks like you didn't initialize it. std::sregex_iterator end;

Galik Over a year ago

@daydayup The end iterator is default constructed by design so that it is equal to iter when iter has run out of matches. Notice at the bottom of the loop I do ++iter. That moves iter on to the next match. When there are no more matches, iter has the same value as the default constructed end. I added another example to show the submatching.

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