Take a look at the following code snippet:
type unionType = (A | B)[]
type unionArr = A[] | B[]
Can you spot any distinctions between these two choices?
Take a look at the following code snippet:
type unionType = (A | B)[]
type unionArr = A[] | B[]
Can you spot any distinctions between these two choices?
There is a distinction between (A | B)[] and A[] | B[]. The former represents an array that can contain elements of type A or type B, while the latter represents an array where all elements are either of type A or all elements are of type B.
class A { a = "a" }
class B { b = "b" }
type unionType = (A | B)[]
let ut: unionType;
ut = [new A(), new A()]
ut = [new B(), new B()]
ut = [new A(), new B()] // mixed array is acceptable
let uti = ut[0] // When reading, we get A | B because any type can come out of the array
ut[0] = new B();
ut[1] = new A();
ut.reduce((s, e) => e instanceof A ? e.a + s : e.b + s, "") // okay
type unionArr = A[] | B[]
let uArr: unionArr;
uArr = [new A(), new A()]
uArr = [new B(), new B()]
uArr = [new A(), new B()] // mixed array is not valid
let uArri = uArr[0] // Reading gives us A | B because we are unsure if we have A[] or B[]
// Type discrepancy: We can assign a B into uArr[0] even if it is holding an A[]
uArr[0] = new B();
uArr[1] = new A();
// error This expression is not callable.
uArr.reduce((s, e) => e instanceof A ? e.a + s : e.b + s, "")
In the case of A[] | B[], the requirement that all array items must be of type A or all of type B is primarily enforced during array assignment, rather than when assigning to individual indexes unfortunately.
Additionally, due to how call signatures work for unions, functions like reduce may become un-callable for A[] | B[], whereas functions like map will function correctly.
Generally, it is recommended to use (A | B)[] unless there are compelling reasons to choose otherwise despite its limitations.
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