What is the reason for Typescript not throwing an error when arguments are missing?

Question

What is the reason for Typescript not throwing an error when arguments are missing?

Here is my code snippet:

type Callback = (a: string, b: string) => void

const observer: Callback = function(a: string): void {
  console.log(a)
}

observer('foo')

I am encountering an issue with Typescript on the last line:

Expected 2 arguments, but got 1.(2554)

I wonder why Typescript only raises an error on the observer call instead of detecting it on line 3 where the observer function is defined?

You can test it here: Test Your Code

typescript

Answer 1

Answer №1

They provide information in their documentation regarding this topic https://www.typescriptlang.org/docs/handbook/type-compatibility.html#comparing-two-functions

In essence, the reason for this behavior is that in JavaScript it is permissible and somewhat common to disregard extra function parameters. For instance, the code below declares a parameter b, but never utilizes it:

const observer: Callback = function(a: string, b: string): void {
  console.log(a);
}

You might agree that this should comply with the Callback type definition. However, what role does b: string play in the argument list? It only impacts the value of observer.length and the return value of observer.toString(). In practical terms, it often serves as redundant typing.

An example of this concept can be observed with higher-order functions like .map, .filter, .reduce on arrays. These functions accept another function as an argument to define behavior. Despite receiving multiple parameters, you may only require one in many scenarios. Therefore, the following scenario should not generate a type error:

const numbers = [1, 2, 3];
const biggerNumbers = numbers.map(value => value + 1);

Even though your function receives a value, index, and array, it is acceptable to ignore the last two (or even all three, although less common).

In summary: The type definition enforces passing specific values into the function, rather than ensuring that the function actually utilizes those values.

Answer 2

They provide information in their documentation regarding this topic https://www.typescriptlang.org/docs/handbook/type-compatibility.html#comparing-two-functions

In essence, the reason for this behavior is that in JavaScript it is permissible and somewhat common to disregard extra function parameters. For instance, the code below declares a parameter b, but never utilizes it:

const observer: Callback = function(a: string, b: string): void {
  console.log(a);
}

You might agree that this should comply with the Callback type definition. However, what role does b: string play in the argument list? It only impacts the value of observer.length and the return value of observer.toString(). In practical terms, it often serves as redundant typing.

An example of this concept can be observed with higher-order functions like .map, .filter, .reduce on arrays. These functions accept another function as an argument to define behavior. Despite receiving multiple parameters, you may only require one in many scenarios. Therefore, the following scenario should not generate a type error:

const numbers = [1, 2, 3];
const biggerNumbers = numbers.map(value => value + 1);

Even though your function receives a value, index, and array, it is acceptable to ignore the last two (or even all three, although less common).

In summary: The type definition enforces passing specific values into the function, rather than ensuring that the function actually utilizes those values.

What is the reason for Typescript not throwing an error when arguments are missing?

Answer №1

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