What is the reason for not applying type guards to callback function arguments?

Question

What is the reason for not applying type guards to callback function arguments?

Check out this TypeScript code snippet for a quick example.

// Setting up the example
declare var data:{info?: {details: string}};
function instant(callback: () => void) {
  callback();
}

// Safeguarding the data
if (data.info) {
  console.log(data.info.details); // ✅ This works!
  instant(() => {
    console.log(data.info.details); // ❌ TypeScript error : 'data.info' may be 'undefined'.(18048)
  });
}

The type guard for data.info ensures it is not null, and infers its type as {details: string}.

But why does TypeScript throw an error in the immediate callback function, warning that data.info might be undefined? What's causing this issue?

typescript callback typeguards

Answer 1

Answer №1

The reason for this occurrence is due to TypeScript's uncertainty about when the callback function will be executed, leading to uncertainty about the validity of type narrowing. Any unknown code may have modified the foo in the interim.

To resolve this, a straightforward approach is to assign the subject of scrutiny to a const. This way, TypeScript acknowledges the immutability and ensures the continued validity of type narrowing.

const temp = foo.bar;
if (temp) {
  console.log(temp.baz);
  immediate(() => {
    console.log(temp.baz);
  });
}

On a side note, TypeScript version 5.4 (currently in beta) has made improvements in handling similar scenarios. However, as confirmed by Jcalz, your specific case may still encounter this issue. For further details, refer to:

Answer 2

The reason for this occurrence is due to TypeScript's uncertainty about when the callback function will be executed, leading to uncertainty about the validity of type narrowing. Any unknown code may have modified the foo in the interim.

To resolve this, a straightforward approach is to assign the subject of scrutiny to a const. This way, TypeScript acknowledges the immutability and ensures the continued validity of type narrowing.

const temp = foo.bar;
if (temp) {
  console.log(temp.baz);
  immediate(() => {
    console.log(temp.baz);
  });
}

On a side note, TypeScript version 5.4 (currently in beta) has made improvements in handling similar scenarios. However, as confirmed by Jcalz, your specific case may still encounter this issue. For further details, refer to:

What is the reason for not applying type guards to callback function arguments?

Answer №1

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