xpressive: How to combine vector of regex to one?

From: Anton Sergunov (a.sergunov_at_[hidden])
Date: 2022-07-08 15:15:08

Hello Boost Users!

I'm seeking some help from the community because can't see answer in
documentation or in source code.

I need to find which regex of N my string matches. I don't like to compare
them one by one, to speed up the process.

It's easy to implement if I have them compile time:
```cpp
sregex re = (s1 = re1)|(s2 = re2)|(s3=re3);
...
std::set<int> foo(const std::string& str) {
 std::set<int> result;
 smatch match;
 if(regex_match(str, match, re)) {
 for(int i = 0; i<3; ++i)
  if(match[i+1]) result.insert(i);
 }
return result;
}
```

But I have a vector:
```cpp
std::set<int> foo(const std::string& str, const std::vector<sregex>&
reVector) {
 sregex re;
 int index = 1;
 for(auto&& item : reVector) {
   re |= (mark_tag(index++) = item); // (1)
 }

 std::set<int> result;
 smatch match;
 if(regex_match(str, match, re)) {
  for(int i = 0; i<reVector; ++i)
    if(match[i+1]) result.insert(i);
  }
 }
return result;
}
```

There at (1) `|=` doesn't work as I expect for some reson, It didn't catch
some values.
If I replace it with
```cpp
 re = re | (mark_tag(index++) = item);
```
it just crashes. If I replace with:
```cpp
 auto t = re;
 re = t | (mark_tag(index++) = item);
```
it making captures of previous tags to be subexpresions matches. So code
doesn't work as I want.

Is it any way to prevent sub-expression capture isolation?

As I understand it should be some intermediate type before sregex to
express expression under construction. So it will not isolate matches of
existing part.

I'm using v1.65.1

https://github.com/boostorg/xpressive/issues/24