TypeScript Basics: Generic Function Inference

• Generics would be tedious if we always had to write them out. Fortunately, they can be inferred when calling generic functions.

• >

  function length<T>(elements: T[]): number {
    return elements.length;
  }
• Note: this code example reuses elements (variables, etc.) defined in earlier examples.

  >

  length([]);

  Result:

  0
• Note: this code example reuses elements (variables, etc.) defined in earlier examples.

  >

  length([1, 2]);

  Result:

  2
• Note: this code example reuses elements (variables, etc.) defined in earlier examples.

  >

  [
    length([1, 2]),
    length(['a']),
  ];

  Result:

  [2, 1]

• We called the function twice without specifying the type parameters. We could've written length<number>(...) and length<string>(...). But it's much nicer to read the code without them!

• How does the compiler do this while preserving type safety?

  • We call length([1, 2]).
  • The compiler sees that length<T> takes an Array<T>.
  • The compiler also sees that [1, 2] is an Array<number>.
  • We passed an Array<number> for the Array<T>, so T must be number.

  Then the whole process repeats for the length(['a']) call. In that case, T is string.

• The two calls to length are checked separately. It's fine that T is number for the first call and string for the second. That's the purpose of generics: to allow us to reuse code in different ways.

• Type safety is maintained throughout all of this. Remember: inference doesn't eliminate type safety!

• Note: this code example reuses elements (variables, etc.) defined in earlier examples.

  >

  const s: string = length([1]);

  Result:

  type error: Type 'number' is not assignable to type 'string'.