To define the provision relation, which determines which properties can be held by which objects.

• §2. Family
• §3. Initial stock
• §4. Typechecking
• §5. Assertion
• §6. Compilation
• §7. Problem message text

§1. Assertions assigning properties ("A scene has a time called the expected duration") generate propositions which assert that a subject, here "scene", provides a property, here "expected duration". The relation needed is:

binary_predicate *R_provision = NULL;

§2. Family. There is just one provision relation, which is the only member of its family.

bp_family *provision_bp_family = NULL;

void ProvisionRelation::start?Usage of ProvisionRelation::start:
Knowledge Module - §3(void) {
    provision_bp_family = BinaryPredicateFamilies::new();
    METHOD_ADD(provision_bp_family, STOCK_BPF_MTID, ProvisionRelation::stock);
    METHOD_ADD(provision_bp_family, TYPECHECK_BPF_MTID, ProvisionRelation::typecheck);
    METHOD_ADD(provision_bp_family, ASSERT_BPF_MTID, ProvisionRelation::assert);
    METHOD_ADD(provision_bp_family, SCHEMA_BPF_MTID, ProvisionRelation::schema);
    METHOD_ADD(provision_bp_family, DESCRIBE_FOR_INDEX_BPF_MTID,
        ProvisionRelation::describe_for_index);
}

§3. Initial stock. There's just one relation of this kind, and it's hard-wired in. The terms are given a very broad range of what they can accept.

void ProvisionRelation::stock?Usage of ProvisionRelation::stock:
§2(bp_family *self, int n) {
    if (n == 1) {
        R_provision =
            BinaryPredicates::make_pair(provision_bp_family,
                BPTerms::new(NULL), BPTerms::new(NULL),
                I"provides", NULL, NULL, NULL,
                PreformUtilities::wording(<relation-names>, PROVISION_RELATION_NAME));
        BinaryPredicates::set_index_details(R_provision, "value", "property");
    }
}

§4. Typechecking. Any property can in principle be assigned to any inference subject, so there's really no restriction on the left term. The right term, of course, has to be a property.

int ProvisionRelation::typecheck?Usage of ProvisionRelation::typecheck:
§2(bp_family *self, binary_predicate *bp,
    kind **kinds_of_terms, kind **kinds_required, tc_problem_kit *tck) {
    if (Kinds::get_construct(kinds_of_terms[1]) == CON_property) return ALWAYS_MATCH;
    Problems::quote_kind(4, kinds_of_terms[1]);
    StandardProblems::tcp_problem(_p_(PM_BadProvides), tck,
        "that asks whether something provides something, and in Inform 'to provide' "
        "means that an object (or value) has a property attached - for instance, "
        "containers provide the property 'carrying capacity'. Here, though, what "
        "comes after 'provides' is %4 rather than the name of a property.");
    return NEVER_MATCH;
}

§5. Assertion. If we assert that, say, vehicles provide "colour", then we are implicitly enabling adjectives formed from its enumerated instances — say, "green" or "blue" — can apply to vehicles, so we must make sure any such meanings are defined.

int ProvisionRelation::assert?Usage of ProvisionRelation::assert:
§2(bp_family *self, binary_predicate *bp,
        inference_subject *infs0, parse_node *spec0,
        inference_subject *infs1, parse_node *spec1) {
    property *prn = Rvalues::to_property(spec1);
    if ((infs0) && (prn)) {
        PropertyPermissions::grant(infs0, prn, TRUE);
        Instances::update_adjectival_forms(prn);
        return TRUE;
    }
    return FALSE;
}

§6. Compilation. Run-time is too late to change which objects provide what, so this relation can't be changed at compile time.

int ProvisionRelation::schema?Usage of ProvisionRelation::schema:
§2(bp_family *self, int task, binary_predicate *bp,
    annotated_i6_schema *asch) {
    if (task == TEST_ATOM_TASK) return RTProperties::test_provision_schema(asch);
    return FALSE;
}

§7. Problem message text. Nothing special is needed here.

void ProvisionRelation::describe_for_index?Usage of ProvisionRelation::describe_for_index:
§2(bp_family *self, OUTPUT_STREAM,
    binary_predicate *bp) {
    WRITE("provision");
}