Circular function reference in Typescript occurs when a function calls itself

Question

Circular function reference in Typescript occurs when a function calls itself

The functionality of this code snippet is rather straightforward; it either returns a function or a string based on an inner function parameter.

function strBuilder(str: string) {
  return function next(str2?: string) {
    if(typeof str2 === "string") {
      return strBuilder(str + str2)
    } else {
      return str
    }
  }
}

Unfortunately, there is a type error that occurs when executing the following expression:

strBuilder("Hello, ")("World")()

I am seeking assistance in properly annotating this function to eliminate the type error, as inference is not providing the desired results.

I have conceptualized a type for this function, but I am unsure how to specify that it should conform to that particular type.

type StringBuilder<S extends string> = (
  str: S
) => (str2?: S) => S extends string ? ReturnType<StringBuilder<S>> : string

Any guidance or support on this matter would be greatly appreciated.

You can find a demo of the issue in the provided playground link here.

typescript recursion functional-programming

Answer 1

Answer №1

To streamline the process, consider defining the return type of strBuilder() as an overloaded function type:

interface StringBuilder {
  (str: string): StringBuilder;
  (): string;
}

function strBuilder(str: string): StringBuilder {
  return function next(str2?: string) {
    if (typeof str2 === "string") {
      return strBuilder(str + str2)
    } else {
      return str
    }
  } as StringBuilder // <-- assert
}

An assertion is necessary here since the compiler cannot fully confirm that function implementations adhere to overloaded call signatures (refer to How to correctly overload functions in TypeScript?).

The code above compiles successfully and yields the expected behavior upon execution:

const str = strBuilder("Hello, ")("World")();
console.log(str.toUpperCase())

You can also opt for a more complex generic conditional approach, although its benefits may vary based on specific scenarios:

interface StringBuilder {
  <S extends string | undefined = undefined>(str?: S):
     S extends string ? StringBuilder : string;
}

This method is akin to your initial implementation, with direct representation of the return type to avoid using ReturnType and potential conflicts with variable names like S. A default value for S ensures functionality when the function is called without arguments.

Similar to the previous approach, a type assertion within strBuilder() is essential due to limitations in verifying generic conditional return types accurately (see microsoft/TypeScript#33912). This doesn't introduce any significant changes regarding type safety.

In essence, both methods deliver the same results for the aforementioned example:

const str = strBuilder("Hello, ")("World")();
console.log(str.toUpperCase())

Unless specific challenges arise, sticking with overloads might be the preferable choice.

Try out the code on the TypeScript Playground

Answer 2

To streamline the process, consider defining the return type of strBuilder() as an overloaded function type:

interface StringBuilder {
  (str: string): StringBuilder;
  (): string;
}

function strBuilder(str: string): StringBuilder {
  return function next(str2?: string) {
    if (typeof str2 === "string") {
      return strBuilder(str + str2)
    } else {
      return str
    }
  } as StringBuilder // <-- assert
}

An assertion is necessary here since the compiler cannot fully confirm that function implementations adhere to overloaded call signatures (refer to How to correctly overload functions in TypeScript?).

The code above compiles successfully and yields the expected behavior upon execution:

const str = strBuilder("Hello, ")("World")();
console.log(str.toUpperCase())

You can also opt for a more complex generic conditional approach, although its benefits may vary based on specific scenarios:

interface StringBuilder {
  <S extends string | undefined = undefined>(str?: S):
     S extends string ? StringBuilder : string;
}

This method is akin to your initial implementation, with direct representation of the return type to avoid using ReturnType and potential conflicts with variable names like S. A default value for S ensures functionality when the function is called without arguments.

Similar to the previous approach, a type assertion within strBuilder() is essential due to limitations in verifying generic conditional return types accurately (see microsoft/TypeScript#33912). This doesn't introduce any significant changes regarding type safety.

In essence, both methods deliver the same results for the aforementioned example:

const str = strBuilder("Hello, ")("World")();
console.log(str.toUpperCase())

Unless specific challenges arise, sticking with overloads might be the preferable choice.

Try out the code on the TypeScript Playground

Circular function reference in Typescript occurs when a function calls itself

Answer №1

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