[m-users.] Why can't `det` be used where `multi` is expected.

[Philip White]

The type [get_opt.option_ops] provides a choice between [option_ops] and
[option_ops_multi], both of which contain short, long, and default
option predicates. The difference is that [option_ops] expects the
predicate for default option values to be [nondet], but
[option_ops_multi] expects it to be [multi].

Below, my predicate, [option_default/2], is declared as [multi], and
this whole program compiles. However, [option_default/2] is clearly
deterministic, and indeed the compiler warns that I could be more
specific. However, if I take the suggestion, then I get nasty
instantiatedness errors. It would seem to me that passing a
deterministic predicate into a place which expects a [multi] predicate
is always valid. What am I not understanding?

```
:- module mds.

:- interface.

:- import_module io.

:- pred main(io.state::di, io.state::uo) is det.

%------------------------------------------------------------------------------%

:- implementation.

:- import_module getopt, bool.

:- pred write_stars(io.state::di, io.state::uo) is det.
write_stars(!IO) :- 
  io.write_string("********************************************************************************\n", !IO).

main(!IO) :- 
  io.command_line_arguments(Argv, !IO),
  OptionOps = option_ops_multi(short_option, long_option, option_default),
  getopt.process_options(OptionOps, Argv, Options, GetoptResult),
  io.write(Argv, !IO),
  io.write(Options, !IO),
  io.write(GetoptResult, !IO),
  io.nl(!IO),
  io.write_string("\n", !IO).

:- type option ---> stars.

:- pred option_default(option::out, option_data::out) is multi.
option_default(stars, bool(yes)).

:- pred short_option(character::in, option::out) is semidet.
short_option('s', stars).

:- pred long_option(string::in, option::out) is semidet.
long_option("stars", stars).
```