ballerina-lang
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ballerina-platform
[Task]: Support `key-constraint` in table semtype
Description
$subject.
In the current nBallerina table semtype implementation key-constraint is not supported. Ref: https://github.com/ballerina-platform/ballerina-lang/pull/42639#discussion_r1582795300
- KS - Key Specifier
- KC - Key Constraint
- KST - KeySemtype
In general,
table<C1> key(KS1)/key<KC1> <: table<C2> key(KS2)/key<KC2> iff
table<C1> <: table<C2> & KeySemType1 <: KeySemType2
Keyless table table<C> can be modeled as C[]. Thefore the keyed-table can be modeled using a tuple [C[], KST]
How do we model KST?
Basically, we encounter 4 scenarios when performing subtype checking.
Sincetable<C> key(KS)/key<KC> <: table<C>,
in all cases, KS = undef and KC=never/undef, KST has to be modeled as ANYDATA(key's top type) or ANY(simpler).
Case 1:
key(f'1, f'2, ...) <: key(f"1, f"2, ...) iff KST1 <: KST2
where KST is either record {|T1 f1; T2 f2; ...; any...;|} or [[T1, "f1"], [T2, "f2"], ..., [any, string]...]
Case 2:
key<KC1> <: key<KC2> iff KST1 <: KST2
where KST = KC. This is straightforward.
Case 3:
key(f1, f2, ...) <: key<KC> iff KST1 <: KST2
KST1 is either record {|T1 f1; T2 f2; ... any...;|} or [[T1, "f1"], [T2, "f2"], ..., [any, string]...]
KST2?
To hold the subtyping KC has to be a tuple where [T1, T2, ...,] is a subtype.
So, if KC is a tuple. KST2 can be modeled as [[T'1, any], [T'2, any], ..., [any, any]...], else KST = KC
Case 4:
key<KC> <: key(f1, f2, ..) iff KST1 <: KST2
KST2 is either record {|T1 f1; T2 f2; ... any...;|} or [[T1, "f1"], [T2, "f2"], ..., [any, string]...]
KST1?
If KC is a tuple it is modeled as [[T1, "f1"], [T2, "f2"], ..., [any, string]...], else KST1=KC
// CASE 1:
type Employee record {|
readonly string name;
readonly string department;
int salary;
|};
type Employee2 record {|
readonly [string, string] name;
int salary;
|};
public function case1() {
table<Employee> key<[string]> t1 = table key(name)[
{name: "John", department: "", salary: 100},
{name: "Jane", department: "",salary: 200}
]; // ALLOWED
table<Employee2> key<[string, string]> t3 = table key(name)[
{name: ["John", ""], salary: 100},
{name: ["Jane", ""],salary: 200}
]; // NOT ALLOWED.
table<Employee2> key<[[string, string]]> t2 = table key(name)[
{name: ["John", ""], salary: 100},
{name: ["Jane", ""],salary: 200}
]; // ALLOWED.
}
// Conclusion: KC being a tuple has special meaning. Tuple fixed member size = no of items in composite key
// CASE 2:
type Employee record {|
readonly string name;
readonly string department;
int salary;
|};
public function case2() {
table<Employee> key<[string]> a = table key(name) [
{name: "John", department: "", salary: 100},
{name: "Jane", department: "",salary: 200}
];
table<Employee> key<string> a1 = a; // NOT ALLOWED. BUG?
table<Employee> key<string> b = table key(name) [
{name: "John", department: "", salary: 100},
{name: "Jane", department: "",salary: 200}
];
table<Employee> key<[string]> b1 = b; // NOT ALLOWED. BUG?
}
// CASE 3:
type Employee record {|
readonly string name;
readonly string department;
int salary;
|};
public function case3() {
table<Employee> key(name, department) a = table key(name, department) [
{name: "John", department: "", salary: 100},
{name: "Jane", department: "", salary: 200}
];
table<Employee> key(name) a1 = a; // NOT ALLOWED. BUG?
table<Employee> key<[string, string]> b = table key(name, department) [
{name: "John", department: "", salary: 100},
{name: "Jane", department: "", salary: 200}
];
table<Employee> key<string> b1 = b; // NOT ALLOWED. BUG?
table<Employee> key<[string]> b2 = b; // NOT ALLOWED. BUG?
table<Employee> key(name) c = table key(name) [
{name: "John", department: "", salary: 100},
{name: "Jane", department: "", salary: 200}
];
table<Employee> key(name, department) c1 = c; // NOT ALLOWED. OK.
table<Employee> key<[string]> d = table key(name) [
{name: "John", department: "", salary: 100},
{name: "Jane", department: "", salary: 200}
];
table<Employee> key<[string, string]> d1 = d; // NOT ALLOWED. OK.
}
CASE III is not a BUG. e.g.
table<Employee> key(name, department) a = table key(name, department) [
{name: "John", department: "IT", salary: 100},
{name: "John", department: "finance", salary: 200}
];
This being assignable to table<Employee> key(name) would not make the name a unique key.
CASE III is not a BUG. e.g.
Based on the above, the new model would be,
- KS - Key Specifier
- KC - Key Constraint
- KST - KeySemtype
In general,
table<C1> key(KS1)/key<KC1> <: table<C2> key(KS2)/key<KC2> iff
table<C1> <: table<C2> & KeySemType1 <: KeySemType2
Keyless table table<C> can be modeled as C[]. Thefore the keyed-table can be modeled using a tuple [C[], KST]
How do we model KST?
Basically, we encounter 4 scenarios when performing subtype checking.
Sincetable<C> key(KS)/key<KC> <: table<C>,
in all cases, KS = undef and KC=never/undef, KST has to be modeled as ANYDATA(key's top type) or ANY(simpler).
Case 1:
key(f'1, f'2, ...) <: key(f"1, f"2, ...) iff KST1 <: KST2
where KST is either record {|T1 f1; T2 f2; ...;|} or [[T1, "f1"], [T2, "f2"], ...]
Case 2:
key<KC1> <: key<KC2> iff KST1 <: KST2
where KST = KC. This is straightforward.
Case 3:
key(f1, f2, ...) <: key<KC> iff KST1 <: KST2
KST1 is either record {|T1 f1; T2 f2; ...; |} or [[T1, "f1"], [T2, "f2"], ...]
KST2?.
If KC is a tuple. KST2 can be modeled as [[T'1, any], [T'2, any], ...], else KST = KC
Case 4:
key<KC> <: key(f1, f2, ..) iff KST1 <: KST2
KST2 is either record {|T1 f1; T2 f2; ...;|} or [[T1, "f1"], [T2, "f2"], ...]
KST1?
If KC is a tuple it is modeled as [[T'1, "f1"], [T'2, "f2"], ....], else KST1=KC