🔎 Search Terms

type narrowing local variable

🕗 Version & Regression Information

I don't see anything related in the FAQ. But I acknowledge that this seem too common to be unknown. Every version seem affected however.

⏯ Playground Link

https://www.typescriptlang.org/play?ts=5.4.2#code/MYewdgzgLgBMCGAbRMC8MAUAzAXJglGgHwwBuIAlgCaGolYb4DcAUC6JLFAKbRoHEYAbxYwxMRN1jw8AVzBVuWCmG5UYAHxhhZAWwBG3AE795i5aqqtxMAPS2YAFQCeAB17AjFV7AjduuhAwUCAwhnAguq6SPIjOMPLwAO7wRtzBABbwXBnppEiy3KLi9jAgWDAABvCVcPBgcFlgAObpKpnpWCDIIEkqzXXIAHTFYgjIGIyCQjDw-ACMTDAAvsyjdg4AggodaQA0MBS+UBTImSoA1pnZVTV1DeBx66WuIBAQFPpxYemVZkoqNSVEY2CgVDBzACEqHQ-wsakIIhsNlKUAyFCCiEBMFasDSACtuMAeFR1jZ4EMQgBlKBeFqMaziZYsZasIA

💻 Code

const call = (f: () => void) => f();

const test = () => {
    let a: undefined | number = undefined;
    // Typescript seems to be completely unaware that the value
    // of `a` can change in the following call.
    call(() => { a = 1; });
    // And there, it still think that `a` can only
    // possibly be `undefined`.
    if (a !== undefined) {
        // this line get rejected
        a.toString();
    }
};

🙁 Actual behavior

It seems that Typescript is not considering a function call to affect local variable. It proceed to narrow the type from this limitation, and thus reject valid code.

Not assigning undefined to a causes the error to disappear BUT this is still a problem, because I might want a default value here (could be a number initially, that later get changed to undefined for example).

For example, and this is worse:

const call = (f: () => void) => f();

const test = () => {
    let a: undefined | number = 42;
    call(() => {
        a = undefined;
    });
    a.toString();
};

This will compile fine, but it will fail at runtime!

🙂 Expected behavior

The type should not be narrowed in this case.

Typescript should detect that a closure is referencing a local variable, and thus, should avoid narrowing its type.

Additional information about the issue

No response