boost regex error: uninitialzed boost::match_results

Question

When running the following

    bool my_compare(const std::string& string_1, const std::string& string_2)
    {
        const boost::regex str_re("so(m)e_(r)e(g)ex");
        boost::smatch str_match;     

        size_t one;
        uint8_t two;
        size_t three;
        if( boost::regex_search(string_1, str_match, str_re) ) {
            one = boost::lexical_cast<size_t>( std::string(str_match[1].first, match[1].second) );
            two = boost::lexical_cast<uint8_t>( std::string(str_match[2].first, match[2].second) );
            three = boost::lexical_cast<size_t>( std::string(str_match[3].first, match[3].second) );
        }
        size_t four;
        uint8_t five;
        size_t six;
        if( boost::regex_search(string_1, str_match, str_re) ) {
            four = boost::lexical_cast<size_t>( std::string(str_match[1].first, match[1].second) );
            five = boost::lexical_cast<uint8_t>( std::string(str_match[2].first, match[2].second) );
            six = boost::lexical_cast<size_t>( std::string(str_match[3].first, match[3].second) );
        }
        return false;
    }

    int main()
    {
       my_compare(some_string1, some_string2);
       return 0;
    }

I'm getting the following error, which I don't understand:

terminate called after throwing an instance of 'boost::exception_detail::clone_impl

' what(): bad lexical cast: source type value could not be interpreted as target

are str_match and match supposed to be the same? It helps to post code that compiles. — Cubbi
– Cubbi, Commented Mar 4, 2014 at 18:41
match[x] is generated by regex. str_match is the object used for the result — zzzbbx
– zzzbbx, Commented Mar 4, 2014 at 18:43
The code shown doesn't initialize or even declare anything called match. — Cubbi
– Cubbi, Commented Mar 4, 2014 at 18:46

Cubbi · Accepted Answer · 2014-03-04 18:52:51Z

1

"so(m)e_(r)e(g)ex"

This regex will populate capture str_match[1] with the string "m", str_match[2] with the string "r" and str_match[3] with the string "g"

 boost::lexical_cast<size_t>( std::string(str_match[1].first, match[1].second) );

This will attempt to convert the string "m" into a value of type size_t, which fails ("m" is not an integer!), and throws the exception

' what(): bad lexical cast: source type value could not be interpreted as target

(by the way, you don't need to jump these hoops with string constructor from a pair of iterators, submatches are directly convertible to strings)

answered Mar 4, 2014 at 18:52

Cubbi

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1 Comment

zzzbbx Over a year ago

the regex was an example which is not what I'm using. However, you helped me find the problem with casting. Thanks!

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