Compiler unable to determine Generic type if not explicitly specified

Here is a simple code snippet that I am working with:

class Model {
    prop1: number;
}

class A<TModel> {
    constructor(p: (model: TModel) => any) {}
    bar = (): A<TModel> => {
        return this;
    }
}

function foo<T>(p: A<Model>) { }

Example 1:

foo(new A(x => x.prop1)) // This works

Example 2:

foo(new A<Model>(x => x.prop1).bar()) // This works

Example 3:

foo(new A(x => x.prop1).bar()) // This doesn't work. (Property 'prop1' does not exist on type '{}'.)

I have encountered an issue where I want example 3 to work in the same way as example 2. It seems that the TypeScript compiler is unable to maintain the generic type if it is not explicitly set and the method "bar" is called immediately after the constructor. My question now is, is this a bug or am I missing something? If I am doing something wrong, how can I resolve this issue?

Please let me know if there are any missing details.

Answer №1

The majority of type inference in TypeScript follows a "forward in time" direction, where it deduces the types of an expression based on the types of its constituents. For instance:

declare const x: string;
declare function f<T>(x: T): Array<T>;
const y = f(x); // y is inferred as Array<string>;

In this scenario, the type of f(x) is established by the type of x and the definition of f. The compiler can comfortably execute this because it simulates the typical behavior of a JavaScript runtime, which involves knowing f and x to generate f(x).


On the other hand, there's also contextual typing, which operates in a somewhat "backwards in time" manner. Here, the compiler is aware of the anticipated type of an expression and partial information about its components, aiming to infer what the missing pieces must have been to produce the expected output. Consider this example:

declare function anyX<T>(): T;
const someX = anyX(); // inferred as {}
declare function f<T>(x: T): Array<T>;
const y: Array<string> = f(anyX()); // anyX() presumed to be string

In this case, since anyX() is generic with no parameters, inferring T directly from its calls is impossible, resulting in {} for someX. However, due to knowledge of y needing to be an Array<string>, the compiler deduces that anyX() should return a string.

While contextual typing is beneficial, it doesn't cover all scenarios. Particularly, inferring the generic type parameter of a generic entity from its properties or methods isn't supported:

type G<T> = { g: T; }
declare function anyG<T>(): G<T>;
const h: string = anyG().g; // error! no contextual typing here

This leads to the issue highlighted in your "Example 3". The compiler computes the return type of

anyG()</code upfront without considering the type of <code>g</code, potentially resulting in errors.</p>

<p>Although the reason behind this lack of contextual typing might stem from performance considerations, adopting manual type parameter specification or wrapping problematic code within functions are viable workarounds:</p>

<pre><code>const h: string = anyG<string>().g; // acceptable

For those encountering recurring issues, leveraging contextual type inference from return type to parameter type through function encapsulation could provide additional safety:

// code snippet provided as the best solution option

These approaches offer solutions when contextual typing falls short. Stay proactive in dealing with such cases effectively. Best of luck!

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