Determine the consistent type for numerous properties

Is it feasible to have Typescript automatically infer the type of multiple properties to be the same, or even infer the types based on a specific property?

For example, consider the following code snippet ->

interface Test<T> {
  value1: T;
  value2: T;
}

const test1:Test<string> = {
    value1: 'string1',
    value2: 'string1'
}

In the above scenario, if we try to assign a number to either value1 or value2, it will result in an error. However, it would be useful if we could omit the generic <string> and let Typescript determine it based on the types of maybe the first property value1.

Initially, I attempted to use a default type for T as 'any' hoping that it would later infer the actual type, but it continues to remain 'any'. As demonstrated below where we end up with a string and a number.

interface Test<T = any> {
  value1: T;
  value2: T;
}

const test1:Test = {
    value1: 'string1',
    value2: 2
}

Answer №1

When using a default generic type parameter, the compiler interprets it as the default type (usually 'any') if not explicitly specified. Unfortunately, TypeScript does not have native support for inferring unspecified type parameters, as requested in this feature request.

To work around this limitation, you can create a utility function that takes a generic input and returns it back:

const asTest = <T>(test: Test<T>) => test;

By calling this function with an object, TypeScript will try to infer the appropriate type for the test parameter, eliminating the need for manual annotations:

// Example usage
const testString = asTest({
  value1: "string1",
  value2: "string2"
})
// const testString: Test<string>

const testStringNumber = asTest({
  value1: "string1",
  value2: 2 // error! 'number' is not assignable to 'string'
})
// const testString: Test<string>

These modifications help achieve the desired behavior.


It's important to note that TypeScript may not always synthesize union types when dealing with unannotated values. In certain cases, like in the example provided, unions of object types are more likely to be inferred by the compiler due to their expected nature.

Another approach, although not directly supported by TypeScript, involves specifying constraints on the generic type parameters to enforce specific type checks:

const asTest = <T, U extends T>(test: { value1: T, value2: U }): Test<T> => test;

This version of asTest uses two generic type parameters where T corresponds to value1 and U must be assignable to

T</code. This prevents the compiler from inferring incorrect union types and enforces strict typing rules.</p>
<p>The code snippet below illustrates how the updated function handles different scenarios:</p>
<pre><code>// Example usage
const testUnion = asTest({
  value1: { a: "" },
  value2: { b: "" } // error! '{ b: string; }' is not assignable to '{ a: string; }'
})

By leveraging constraints on the generic type parameters, you can ensure accurate type inference and prevent unwanted unions in your TypeScript code.


View the Playground link to experiment with the code further.

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