Generic Inference | Engineering Interview Kit

TL;DR

When you call a Generic function, you usually don’t need to explicitly pass the type inside angle brackets (<string>). TypeScript’s Generic Inference looks at the arguments you pass in and automatically figures out what T should be.

Mental Model

How It Works

TypeScript is extremely smart at pattern matching. If your function signature is function makeArray<T>(item: T): T[], and you call makeArray("hello"), TS looks at the first argument, sees a string, matches it against item: T, and instantly decides that T = string.

This keeps your code clean and removes visual clutter while maintaining 100% type safety.

Example

function merge<U, V>(obj1: U, obj2: V): U & V {
    return { ...obj1, ...obj2 };
}

// We DO NOT need to write: merge<{name: string}, {age: number}>(...)
// TS infers U and V automatically based on the two objects passed in.
const mergedObj = merge(
    { name: "Alice" }, 
    { age: 30 }
);

// mergedObj is perfectly typed as: { name: string } & { age: number }
console.log(mergedObj.name);
console.log(mergedObj.age);

Common Interview Questions

When does Generic Inference fail?

Inference fails (or infers unknown / any) when there are no arguments from which TypeScript can deduce the type. For example, if you have function fetchApi<T>(): Promise<T>, calling fetchApi() gives TS zero clues about what T is. In this case, you must explicitly provide the type: fetchApi<User>().

Can you provide a default type for a Generic?

Yes! Just like default function parameters, you can provide default generic parameters: <T = string>. If TS cannot infer the type, and you don’t provide one, it will fall back to string.