如何匹配以“！”开头的十进制数字 在C ++中

Question

How do I match all the decimal numbers starting with "!" (bang) in the given string? I have written the following code but it fails with an assert

#include<iostream>
#include<regex>

int main()
{
    std::string s1("{!112,2,3}");
    std::regex e(R"(\!\d+)", std::regex::grep);

    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;
    }
}

assert

terminate called after throwing an instance of 'std::regex_error'
  what():  regex_error
Aborted (core dumped)

Answer 1

First, make sure you are using the latest GCC compiler.

Then use R"(!(\\d+))" pattern that matches an exclamation mark and then captures one or more digits Iinto Group 1.

Then just grab (*iter)[1] that holds your values while iterating the matches.

See the C++ demo :

#include<iostream>
#include<regex>
int main() {
   std::string s1("{!112,2,3} {!346,765,8}"); 
   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 << "Value: " << (*iter)[1] << std::endl; 
       ++iter; 
    }
}

Output:

{!112,2,3} {!346,765,8} 
Value: 112
Value: 346