Issue
I am experiencing some weird behaviour in TypeScript (v4.6.2). Why does (a) work but (b) and (c) don't?
const a: string[] | null = []
if (a?.length > 1) {
console.log(1)
}
const b = [] as (string[] | null)
if (b?.length > 1) {
console.log(1)
}
const c: string[] | null = [] as (string[] | null)
if (c?.length > 1) {
console.log(1)
}
Solution
Consider the following:
const a: string[] | null = []
a // string[]
// TS has narrowed this to the array type in the union
// because you assigned an array to the variable after the type annotation
if (a?.length > 1) { // ok
console.log(1)
}
const b = [] as (string[] | null)
b // string[] | null
// You asserted that the array could be null,
// so it is now possibly null in the type system
if (b?.length > 1) {
// ~~~~~~~~~
// Object is possibly 'undefined'.(2532)
// TS won't allow comparison of a number to undefined
console.log(1)
}
const c: string[] | null = [] as (string[] | null)
c // string[] | null
// Another case of assertion
if (c?.length > 1) {
// ~~~~~~~~~
// Object is possibly 'undefined'.(2532)
// TS won't allow comparison of a number to undefined
console.log(1)
}
// I suggest one of these patterns instead:
declare const maybeNullMaybeArray: string[] | null;
if ((maybeNullMaybeArray?.length ?? 0) > 1) {
// ^^^^
// Use nullish coalescing operator to evaluate the expression to 0
// in the case that it is nullish, so that the comparison will
// always be between two numbers
maybeNullMaybeArray // string[] | null
// but the type is still not narrowed, because the true condition
// could have resulted from the evaluation to the literal zero
// instead of a value from the `length` property on the variable
// so, alternatively:
}
if (maybeNullMaybeArray && (maybeNullMaybeArray.length > 1)) {
// ^^^^^^^^^^^^^^^^^^^^^^^^^^
// The array definitely exists at this point
// so if we got this far, we can use it as an array:
maybeNullMaybeArray // string[]
}
See:
type assertions in the TS handbook
Answered By - jsejcksn

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