Type relations
Type hierarchy
Model with properties
When determining if type S
can be assigned to type T
, if T
is a model with properties, it checks whether all those properties are present in S
and if their types can be assigned to the type of the corresponding property in T
.
For instance,
model T {
foo: string;
bar: int32;
}
// Valid
model S { // When properties types are the exact same
foo: string;
bar: int32;
}
model S { // When the properties types are literal assignable to the target type
foo: "abc";
bar: 123;
}
model S {
foo: string;
bar: int8; // int8 is assignable to int16
}
model S {
foo: string;
bar: int32;
otherProp: boolean; // Additional properties are valid.
}
// Invalid
model S { // Missing property bar
foo: string;
}
model S {
foo: string;
bar: int64; // int64 is NOT assignable to int32
}
Record<T>
A record is a model indexed with a string with a value of T. It represents a model where all properties (string keys) are assignable to the type T. You can assign a model expression where all the properties are of type T or another model that is
also a Record<T>
.
// Represent an object where all the values are int32.
alias T = Record<int32>;
// Valid
alias S = {
foo: 123;
bar: 345;
};
alias S = {
foo: int8;
bar: int32;
};
model S is Record<int32>;
model S is Record<int32> {
foo: 123;
}
// Invalid
alias S = {
foo: "abc";
bar: 456;
};
alias S = {
foo: int64;
bar: int32;
};
model S {
foo: 123;
bar: 456;
}
Why isn't the last case assignable to Record<int32>
?
In this scenario,
alias T = Record<int32>;
model S {
foo: 123;
bar: 456;
}
The reason why model S
is not assignable, but the model expression { foo: 123; bar: 456; }
is, is because model S could be extended with additional properties that might not be compatible.
For instance, if you add a new model,
model Foo is S {
otherProp: string;
}
Here, Foo
is assignable to S
following the model with property logic, and if S
was assignable to Record<int32>
, Foo
would also be passable. However, this is now invalid as otherProp
is not an int32
property.