std::regex_iterator::operator++, operator++(int)

From cppreference.com
regex_iterator& operator++();
(since C++11)
regex_iterator operator++(int);
(since C++11)

Advances the iterator on the next match.

At first, a local variable of type BidirIt is constructed with the value of match[0].second.

If the iterator holds a zero-length match and start == end, *this is set to end-of-sequence iterator and the function returns.

Otherwise, if the iterator holds a zero-length match the operator invokes the following:

regex_search(start, end, match, *pregex,
                  flags | regex_constants::match_not_null |
                          regex_constants::match_continuous);

If the call returns true, the function returns.

Otherwise the operator increments start and continues as if the most recent match was not a zero-length match.

If the most recent match was not a zero-length match, the operator sets flags to flags | regex_constants::match_prev_avail and invokes the following:

regex_search(start, end, match, *pregex, flags);

If the call returns false, the iterator sets *this to the end-of-sequence iterator, the function returns.

In all cases in which the call to regex_search returns true, match.prefix().first will be equal to the previous value of match[0].second and for each index i in the range [0, match.size()) for which match[i].matched is true, match[i].position() will return distance(begin, match[i].first).

This means that match[i].position() gives the offset from the beginning of the target sequence, which is often not the same as the offset from the sequence passed in the call to regex_search.

It is unspecified how the implementation makes these adjustments. This means that a compiler may call an implementation-specific search function, in which case a user-defined specialization of regex_search will not be called.

The behavior is undefined if the iterator is end-of-sequence iterator.

[edit] Parameters

(none)

[edit] Return value

1) *this
2) The previous value of the iterator.