When attempting to utilize a generic type within a TypeScript function:
const func: <T extends number>() => void = () => {
const x: T = 1
}
An error message is generated:
Cannot find name 'T'. TS2304
69 | const func: <T extends number>() => void = () => {
> 70 | const x: T = 1
| ^
71 | }
Is there a way to effectively utilize generic types within a function instead of solely within its signature?
For those interested in utilizing arrow functions, consider applying the inferred typing approach discussed in this response.
const func = <T extends number>(x: T) => x;
By utilizing the alternative function notation, we can eliminate that particular error:
function customFunction<U extends string>(): void {
const y: U = 'hello'
}
As a result, we encounter a more logical error message:
Type ''hello'' is not assignable to type 'U'.
''hello'' is assignable to the constraint of type 'U', however, 'U' might be instantiated with a different subtype of constraint 'string'. TS2322
69 | export function func<T extends number>(): void {
> 70 | const x: T = 1
| ^
71 | }
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