What is the reason behind a TypeScript compiler error being triggered by an 'if-else' statement, while a ternary operator construct that appears identical does not raise any errors?

Question

What is the reason behind a TypeScript compiler error being triggered by an 'if-else' statement, while a ternary operator construct that appears identical does not raise any errors?

My function is designed to either return a value of IDBValidKey or something that has been converted to IDBValidKey. When I use the ternary operator to write the function, it works fine. However, if I try to write it as an if-else statement, it causes a compiler error:

interface IDBValidKeyConvertible<TConverted extends IDBValidKey> {
    convertToIDBValidKey: () => TConverted;
}

function isIDBValidKeyConvertible<TConvertedDBValidKey extends IDBValidKey>(object: unknown): object is IDBValidKeyConvertible<TConvertedDBValidKey> {
    return typeof((object as IDBValidKeyConvertible<TConvertedDBValidKey>).convertToIDBValidKey) === "function";
}

type IDBValidKeyOrConverted<TKey> = TKey extends IDBValidKeyConvertible<infer TConvertedKey> ? TConvertedKey : TKey;

function getKeyOrConvertedKey<TKey extends IDBValidKey | IDBValidKeyConvertible<any>>(input: TKey): IDBValidKeyOrConverted<TKey> {
    if (isIDBValidKeyConvertible<IDBValidKeyOrConverted<TKey>>(input)) {
        return input.convertToIDBValidKey();
    } else {
        return input;
    }
}

function getKeyOrConvertedKeyTernary<TKey extends IDBValidKey | IDBValidKeyConvertible<any>>(input: TKey): IDBValidKeyOrConverted<TKey> {
    return (isIDBValidKeyConvertible<IDBValidKeyOrConverted<TKey>>(input)) ? input.convertToIDBValidKey() : input;
}

getKeyOrConvertedKeyTernary does not produce any errors. However, the else block of getKeyOrConvertedKey results in this error:

Type 'TKey' is not assignable to type 'IDBValidKeyOrConverted<TKey>'.
  Type 'string | number | Date | ArrayBufferView | ArrayBuffer | IDBArrayKey | IDBValidKeyConvertible<any>' is not assignable to type 'IDBValidKeyOrConverted<TKey>'.
    Type 'string' is not assignable to type 'IDBValidKeyOrConverted<TKey>'.

Shouldn't the ternary operator and the if-else statement be equivalent in this context?

Thank you!

typescript typescript-generics

Answer 1

Answer №1

What is the reason behind the TypeScript compiler error when using an 'if-else' statement compared to a ternary operator construct?

Short Answer

The TypeScript compiler interprets an if-else statement as having multiple independent types within each expression, while a ternary operator is seen as an expression with a union type of its true and false sides. In certain cases, the type union in the ternary construct may not trigger a compiler error.

Detailed Answer

Are the ternary operator and if-else statement truly equivalent?

Not exactly.

The distinction lies in the ternary operator being an expression. Refer to a discussion by Ryan Cavanaugh on the topic, where he explains the difference between a ternary operator and an if/else statement. Essentially, the type of a ternary expression is a combination of its true and false outcomes.

In your specific scenario, the type of your ternary expression is any, which is why the compiler does not raise an issue. Your ternary operation involves a union between the input type and the return type of input.convert(). Since, at compile time, the input type extends Container<any>, the return type of input.convert() also becomes any. Consequently, a union with any culminates in the entire ternary expression being of type any.

One quick fix for you is to substitute any with unknown in

<TKey extends IDBValidKey | IDBValidKeyConvertible<any>

. This adjustment will prompt both the if-else statement and the ternary operator to trigger a compiler error.

Simplified Example

Explore this playground link featuring a simplified version of your query. Experiment with changing any to unknown to witness the compiler's varying responses.

interface Container<TValue> {
  value: TValue;
}

declare function hasValue<TResult>(
  object: unknown
): object is Container<TResult>;

// Change any to unknown.
const funcIfElse = <T extends Container<any>>(input: T): string => {
  if (hasValue<string>(input)) {
    return input.value;
  }

  return input;
};

// Change any to unknown.
const funcTernary = <T extends Container<any>>(input: T): string =>
  hasValue<string>(input)
    ? input.value
    : input;

Answer 2

What is the reason behind the TypeScript compiler error when using an 'if-else' statement compared to a ternary operator construct?

Short Answer

The TypeScript compiler interprets an if-else statement as having multiple independent types within each expression, while a ternary operator is seen as an expression with a union type of its true and false sides. In certain cases, the type union in the ternary construct may not trigger a compiler error.

Detailed Answer

Are the ternary operator and if-else statement truly equivalent?

Not exactly.

The distinction lies in the ternary operator being an expression. Refer to a discussion by Ryan Cavanaugh on the topic, where he explains the difference between a ternary operator and an if/else statement. Essentially, the type of a ternary expression is a combination of its true and false outcomes.

In your specific scenario, the type of your ternary expression is any, which is why the compiler does not raise an issue. Your ternary operation involves a union between the input type and the return type of input.convert(). Since, at compile time, the input type extends Container<any>, the return type of input.convert() also becomes any. Consequently, a union with any culminates in the entire ternary expression being of type any.

One quick fix for you is to substitute any with unknown in

<TKey extends IDBValidKey | IDBValidKeyConvertible<any>

. This adjustment will prompt both the if-else statement and the ternary operator to trigger a compiler error.

Simplified Example

Explore this playground link featuring a simplified version of your query. Experiment with changing any to unknown to witness the compiler's varying responses.

interface Container<TValue> {
  value: TValue;
}

declare function hasValue<TResult>(
  object: unknown
): object is Container<TResult>;

// Change any to unknown.
const funcIfElse = <T extends Container<any>>(input: T): string => {
  if (hasValue<string>(input)) {
    return input.value;
  }

  return input;
};

// Change any to unknown.
const funcTernary = <T extends Container<any>>(input: T): string =>
  hasValue<string>(input)
    ? input.value
    : input;

Answer 3

Answer №2

There are two distinct issues happening concurrently:

There appears to be a bug or constraint in Typescript, but it is contrary to what your query assumes.

Interestingly, using the ternary operator resolves the issue.

The error displayed for the if-else statement is accurate, and the ternary operator should also trigger an error.

It's common to believe our code is flawless, which can lead to overlooking the actual problem at hand - thus pointing to the second issue:

Your code lacks logical reasoning. (Apologies for being direct.)

With the simplified version of the code presented in the question, issue #2 should be more apparent. I plan to provide a comprehensive explanation when time permits.

[1] It's plausible that this challenge might not be a Typescript anomaly, but rather anticipated behavior stemming from subtle differences between if-else and ? : that are unfamiliar to me. Given Javascript's history of idiosyncrasies, this wouldn't be surprising. [EDIT: Refer to Shaun Luttin's response which coincided with my own.]

Answer 4

There are two distinct issues happening concurrently:

There appears to be a bug or constraint in Typescript, but it is contrary to what your query assumes.

Interestingly, using the ternary operator resolves the issue.

The error displayed for the if-else statement is accurate, and the ternary operator should also trigger an error.

It's common to believe our code is flawless, which can lead to overlooking the actual problem at hand - thus pointing to the second issue:

Your code lacks logical reasoning. (Apologies for being direct.)

With the simplified version of the code presented in the question, issue #2 should be more apparent. I plan to provide a comprehensive explanation when time permits.

[1] It's plausible that this challenge might not be a Typescript anomaly, but rather anticipated behavior stemming from subtle differences between if-else and ? : that are unfamiliar to me. Given Javascript's history of idiosyncrasies, this wouldn't be surprising. [EDIT: Refer to Shaun Luttin's response which coincided with my own.]

What is the reason behind a TypeScript compiler error being triggered by an 'if-else' statement, while a ternary operator construct that appears identical does not raise any errors?

Answer №1

Short Answer

Detailed Answer

Simplified Example

Answer №2

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