Issues arise when attempting to enforce type-safety in TypeScript while using the JSON.parse

Question

Issues arise when attempting to enforce type-safety in TypeScript while using the JSON.parse

Is it possible that type-safety is compromised in TypeScript when dealing with JSON parsing?

I should be triggering an error, but I'm not:

interface Person {
  name: string
}

const person: Person = somePossibleFalsey ? JSON.parse(db.person) : undefined

In the above scenario, there should be a type check failure, but surprisingly, there isn't. The db.person may be missing, resulting in person being set to undefined. However, Person should never be equal to undefined. This anomaly seems to be connected to the usage of JSON.parse.

To confirm the expected error output, here is another code snippet which correctly flags an error:

const person: Person = Math.random() > .5 ? { name: 'Arthur' } : undefined

The aforementioned code triggers the appropriate TypeScript error message:

Type '{ name: string; } | undefined' is not assignable to type 'Person'.
  Type 'undefined' is not assignable to type 'Person'.ts(2322)

Why does JSON.parse seem to bypass type-safety measures? Is there another underlying issue at play here?

javascript json typescript type-safety

Answer 1

Answer №1

After using JSON.parse, you will get an any result. Since any can include any type, including undefined, using any | undefined is essentially the same as just using any.

To correctly type the output of JSON.parse, use as like this:

const person: Person = db.person ? JSON.parse(db.person) as Person : undefined; 
// Error: Type 'Person | undefined' is not assignable to type 'Person'. Type 'undefined' is not assignable to type 'Person'

NOTE: There may be confusion about the any type. This disables all type safety measures, allowing every type to be assigned to any and vice versa. For example:

const myAnyFunc = (): any => {return undefined;};
const myPerson: Person = myAnyFunc();

When this occurs, there is no TypeError thrown. It's not unique to JSON.parse(), but rather a challenge with anything that returns any. For more information on any, check out this helpful TypeScript book.

Answer 2

After using JSON.parse, you will get an any result. Since any can include any type, including undefined, using any | undefined is essentially the same as just using any.

To correctly type the output of JSON.parse, use as like this:

const person: Person = db.person ? JSON.parse(db.person) as Person : undefined; 
// Error: Type 'Person | undefined' is not assignable to type 'Person'. Type 'undefined' is not assignable to type 'Person'

NOTE: There may be confusion about the any type. This disables all type safety measures, allowing every type to be assigned to any and vice versa. For example:

const myAnyFunc = (): any => {return undefined;};
const myPerson: Person = myAnyFunc();

When this occurs, there is no TypeError thrown. It's not unique to JSON.parse(), but rather a challenge with anything that returns any. For more information on any, check out this helpful TypeScript book.

Answer 3

Answer №2

The main question at the heart of this issue is why the ternary expression allows undefined as one of the possible values to be assigned to Person. A discussion on a Github thread explains that this behavior is intentional. The ternary expression combines the first return type with the second, resulting in any | undefined, which ultimately collapses to just any, making it a valid assignment. This is due to any being the weakest type in TypeScript.

let n: number;

let x: any = {}
n = x.z ? x.z : undefined // no error

let y: any = {}
if (y.z) {
    n = y.z
} else {
    n = undefined // error
}

(source)

An interesting observation made here is that the inconsistency arises from how the compiler interprets the ternary expression versus an if-else block. In the former, it sees a single return of type any, while in the latter, there are two return sites causing an error when assigning undefined.

The complexity lies in the nature of ternaries as expressions needing a single type. The current implementation infers a type of any | undefined which is inherently unsound given the nature of any. Changing this behavior would pose challenges in scenarios like nested statements or multiple ternaries within a statement.

To properly check types as suggested, the compiler would have to individually evaluate each combination of true/false for every ternary occurrence, leading to a significant increase in complexity and potential errors.

It seems that the existing behavior, flawed as it may be, is the most practical solution at this point.

Answer 4

The main question at the heart of this issue is why the ternary expression allows undefined as one of the possible values to be assigned to Person. A discussion on a Github thread explains that this behavior is intentional. The ternary expression combines the first return type with the second, resulting in any | undefined, which ultimately collapses to just any, making it a valid assignment. This is due to any being the weakest type in TypeScript.

let n: number;

let x: any = {}
n = x.z ? x.z : undefined // no error

let y: any = {}
if (y.z) {
    n = y.z
} else {
    n = undefined // error
}

(source)

An interesting observation made here is that the inconsistency arises from how the compiler interprets the ternary expression versus an if-else block. In the former, it sees a single return of type any, while in the latter, there are two return sites causing an error when assigning undefined.

The complexity lies in the nature of ternaries as expressions needing a single type. The current implementation infers a type of any | undefined which is inherently unsound given the nature of any. Changing this behavior would pose challenges in scenarios like nested statements or multiple ternaries within a statement.

To properly check types as suggested, the compiler would have to individually evaluate each combination of true/false for every ternary occurrence, leading to a significant increase in complexity and potential errors.

It seems that the existing behavior, flawed as it may be, is the most practical solution at this point.

Issues arise when attempting to enforce type-safety in TypeScript while using the JSON.parse

Answer №1

Answer №2

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