Is there a way to determine the return type without using a type variable?

I am striving to develop a function that can clone an array while maintaining the original data types. The current issue I am facing is that TypeScript is declaring the return type as any:

export function cloneArray(inputArray) {
    return [...inputArray];
}
// result will have any[] as type
const result = cloneArray([1, 2, 3]);

Is there a way to instruct TypeScript to set the return type of cloneArray to be the same as the input type, such as an array of numbers in this scenario?

Potential Solution

I am aware that I can achieve this using a type variable:

export function cloneArray<T>(inputArray): T[] {
    return [...inputArray];
}
// result will have number[] as type (only because I specified T as number)
const result = cloneArray<number>([1, 2, 3]);

I am looking for a way to avoid the workaround mentioned above, as it requires me to define the type in advance.

typescript

Answer №1

When you specify the argument as something that relies on T (like T[] in this case), the generic will be automatically inferred from the argument, eliminating the need to explicitly declare it when calling the function:

function exampleFunction<T>(inputArray: T[]): T[] {
  return [...inputArray];
}

const result = exampleFunction([1, 2, 3]); // The variable `result` will have the type `number[]`

Answer №2

Although you may claim to prefer not using a generic type, it is actually the correct approach. However, your current declaration could benefit from utilizing it more effectively:

export function someFn<T>(inputArray: T[]): T[] {
    return [...inputArray];
}

const numberResult: number[] = someFn([1, 2, 3]);       //correct
const stringResult: string[] = someFn(["a", "b", "c"]); //correct
const mixedResult: string[] = someFn(["a", 2]);         //error

Playground for TypeScript

If you specify that the output will be an array of the same type as the input, there is no need to explicitly define the generic type argument.

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