The following code snippet illustrates that in the else statement, it is evident that b cannot be undefined (a||b returns truthy and since a is falsy, b must be truthy).
Why does Typescript show the error 'b' is possibly 'undefined', and what steps can I take to prevent this issue?
function test(a: string | undefined, b: string | undefined) {
if (a || b) {
if (a) {
const x = a.length
} else {
const y = b.length
}
}
}
To troubleshoot this issue, you can follow the steps below:
function test(x: string | undefined, y: string | undefined) {
type X1 = typeof x; // string | undefined
type Y1 = typeof y; // string | undefined
if (x || y) {
type X2 = typeof x; // string | undefined
type Y2 = typeof y; // string | undefined
if (x) {
type X3 = typeof x; // string
type Y3 = typeof y; // string | undefined
const a = x.length
} else {
type X4 = typeof x; // string | undefined
type Y4 = typeof y; // string | undefined
const b = y.length
}
}
}
In essence, (x || y) is unable to narrow down the types.
The conditional flow analysis checks for !== undefined and !== null conditions (a documented limitation of TS), so it's recommended to rewrite it like this:
function test(a: string | undefined, b: string | undefined) {
if (a !== undefined ) {
const aLength = a.length; // number
} else if(b !== undefined){
const bLength = b.length; // number
}
}
Both x and y are declared as number | undefined in Typescript, indicating that they can either have a number value or be undefined. If you are confident that the function will always receive a value, you can use the ! (non-null assertion operator) to access their properties without checking for null or undefined.
const result = x!.toFixed(2);
To learn more about the non-null assertion operator, check out: non-null assertion operator
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