Why do optional values of objects remain optional when being checked in an if statement in Typescript?

Question

Why do optional values of objects remain optional when being checked in an if statement in Typescript?

Recently at work, I encountered this code snippet and was left wondering why Typescript couldn't comprehend the logic behind it. If 'test' in the object can either be undefined or a string, shouldn't it logically infer that within an if statement, 'test' must be treated as a string?

interface IObj {
  test?: string;
}

const obj: IObj = {
  test: "why does this not work?",
};

if (obj.test) {
  const { test } = obj;

  () => testFunc(obj.test); // error (why does this not work?)
  () => testFunc(test); // works 
}

function testFunc(testString: string) {
  return testString;
}

I define an object (obj) with an optional property 'test'.
I then proceed to verify the value of 'test' within an if condition.

The interesting thing is that destructuring works perfectly fine, except if you attempt to utilize the object and its value simultaneously.

typescript object if-statement

Answer 1

Answer №1

An attribute marked with ? (such as 'test' in IObj) has the potential of being undefined, leading to the type of test being string | undefined.

When you validate the existence of the test attribute in your if statement, you are explicitly stating that test must not be undefined in order to proceed into the block. Referencing obj.test or the newly defined variable test should yield a string value.

The issue arises when you utilize obj.test within an arrow function, creating a new scope and negating the validation performed earlier. Essentially, TypeScript cannot assure that obj.test will remain defined when the function is invoked.

However, this uncertainty is avoided with the test variable because it is instantiated within the if statement where obj.test is guaranteed to be a string, hence making test a string without any possibility of being undefined.

Let's consider the following scenario:

if (obj.test) {
  const { test } = obj;

  setTimeout(() => {
    testFunc(obj.test);
    testFunc(test);
  }, 5000);

  obj.test = undefined;
}

setTimeout introduces another scope and, being asynchronous, it executes the function after X milliseconds (in this case, 5 seconds). As it does not block the execution flow, the code reaches obj.test = undefined prior to executing testFunc(obj.test) inside the setTimeout. This serves as an example of a potential issue that may occur.

Answer 2

An attribute marked with ? (such as 'test' in IObj) has the potential of being undefined, leading to the type of test being string | undefined.

When you validate the existence of the test attribute in your if statement, you are explicitly stating that test must not be undefined in order to proceed into the block. Referencing obj.test or the newly defined variable test should yield a string value.

The issue arises when you utilize obj.test within an arrow function, creating a new scope and negating the validation performed earlier. Essentially, TypeScript cannot assure that obj.test will remain defined when the function is invoked.

However, this uncertainty is avoided with the test variable because it is instantiated within the if statement where obj.test is guaranteed to be a string, hence making test a string without any possibility of being undefined.

Let's consider the following scenario:

if (obj.test) {
  const { test } = obj;

  setTimeout(() => {
    testFunc(obj.test);
    testFunc(test);
  }, 5000);

  obj.test = undefined;
}

setTimeout introduces another scope and, being asynchronous, it executes the function after X milliseconds (in this case, 5 seconds). As it does not block the execution flow, the code reaches obj.test = undefined prior to executing testFunc(obj.test) inside the setTimeout. This serves as an example of a potential issue that may occur.

Answer 3

Answer №2

Ever since the release of typescript 4.9, you have the option to utilize the latest satisfies operator instead of traditional type-assertion:

interface IObj {
  test?: string;
}

const obj = {
  test: "why does this not work?",
} satisfies IObJ;

if (obj.test) {
  const { test } = obj;

  () => testFunc(obj.test); // performs as expected
  () => testFunc(test); // functions correctly
}

function testFunc(testString: string) {
    console.log(testString);
  return testString;
}

Playground Example

Answer 4

Ever since the release of typescript 4.9, you have the option to utilize the latest satisfies operator instead of traditional type-assertion:

interface IObj {
  test?: string;
}

const obj = {
  test: "why does this not work?",
} satisfies IObJ;

if (obj.test) {
  const { test } = obj;

  () => testFunc(obj.test); // performs as expected
  () => testFunc(test); // functions correctly
}

function testFunc(testString: string) {
    console.log(testString);
  return testString;
}

Playground Example

Why do optional values of objects remain optional when being checked in an if statement in Typescript?

Answer №1

Answer №2

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