使用boost :: regex_replace的条件替换字符串

Question

I want to simplify the signs in a mathematical expression using regex_replace, here is a sample code:

string entry="6+-3++5";
boost::regex signs("[\-\+]+");
cout<<boost::regex_replace(entry,signs,"?")<<endl;

The output is then 6?3?5. My question is: How can I get the proper result of 6-3+5 with some neat regular expression tools? Thanks a lot.

Tried something else with sregex_iterator and smatch, but still has some problem:

string s="63--17--42+5555";
collect_sign(s);
Output is 
63+17--42+5555+42+5555+5555
i.e.
63+(17--42+5555)+(42+5555)+5555

It seems to me that the problem is related to the match.suffix(), Could anybody help please? The collect_sign function basically just iterate through every sign strings, convert it to "-"/"+" if the number of "-" is odd/even, and then stitch together the suffix expression of the signs.

void collect_sign(string& entry)
{ 
    boost::regex signs("[\-\+]+");
    string output="";
    auto signs_begin = boost::sregex_iterator(entry.begin(), entry.end(), signs);
    auto signs_end = boost::sregex_iterator();
    for (boost::sregex_iterator it = signs_begin; it != signs_end; ++it) 
    {
        boost::smatch match = *it;
        if (it ==signs_begin)
            output+=match.prefix().str();
        string match_signs = match.str();
        int n_minus=count(match_signs.begin(),match_signs.end(),'-');
        if (n_minus%2==0)
            output+="+";
        else
            output+="-";
        output+=match.suffix();
    }
    cout<<"simplify to: "<<output<<endl;
}

Answer 1

Use:

[+\-*\/]*([+\-*\/])

Replace:

$1

You can test here

Answer 2

If you just want a mathematical simplification, you can use:

s = boost::regex_replace(s, boost::regex("(?:++|--"), "+", boost::format_all);
s = boost::regex_replace(s, boost::regex("(?:+-|-+"), "-", boost::format_all);