
[;1m  inspect(MP, Item)[0m

[;;4mSince[0m:
  OTP 17.0

  Takes a compiled regular expression and an item, and returns the
  relevant data from the regular expression.

  The only supported item is [;;4mnamelist[0m, which returns the tuple [;;4m[0m
  [;;4m{namelist, [binary()]}[0m, containing the names of all (unique)
  named subpatterns in the regular expression.

  For example:

    1> {ok,MP} = re:compile("(?<A>A)|(?<B>B)|(?<C>C)").
    {ok,{re_pattern,3,0,0,
                    <<69,82,67,80,119,0,0,0,0,0,0,0,1,0,0,0,255,255,255,255,
                      255,255,...>>}}
    2> re:inspect(MP,namelist).
    {namelist,[<<"A">>,<<"B">>,<<"C">>]}
    3> {ok,MPD} = re:compile("(?<C>A)|(?<B>B)|(?<C>C)",[dupnames]).
    {ok,{re_pattern,3,0,0,
                    <<69,82,67,80,119,0,0,0,0,0,8,0,1,0,0,0,255,255,255,255,
                      255,255,...>>}}
    4> re:inspect(MPD,namelist).
    {namelist,[<<"B">>,<<"C">>]}

  Notice in the second example that the duplicate name only occurs
  once in the returned list, and that the list is in alphabetical
  order regardless of where the names are positioned in the regular
  expression. The order of the names is the same as the order of
  captured subexpressions if [;;4m{capture, all_names}[0m is specified as
  an option to [;;4mrun/3[0m. You can therefore create a name-to-value
  mapping from the result of [;;4mrun/3[0m like this:

    1> {ok,MP} = re:compile("(?<A>A)|(?<B>B)|(?<C>C)").
    {ok,{re_pattern,3,0,0,
                    <<69,82,67,80,119,0,0,0,0,0,0,0,1,0,0,0,255,255,255,255,
                      255,255,...>>}}
    2> {namelist, N} = re:inspect(MP,namelist).
    {namelist,[<<"A">>,<<"B">>,<<"C">>]}
    3> {match,L} = re:run("AA",MP,[{capture,all_names,binary}]).
    {match,[<<"A">>,<<>>,<<>>]}
    4> NameMap = lists:zip(N,L).
    [{<<"A">>,<<"A">>},{<<"B">>,<<>>},{<<"C">>,<<>>}]
