What impact does control flow have on narrowing variable types?

Having reviewed this particular question, my current focus is on understanding how variables can be narrowed down using control flow. An example:

type T1 = {
  a?: {
    b: string
  }
}

const t1: T1 = {}

t1.a.b // displays error, possibly undefined
t1.a = { b: "foo" }
t1.a.b // now works as expected, TypeScript recognizes property a exists

The specific scenario I am struggling with at the moment:

const t1: T1 = {}
t1.a = { b: "foo" }

function foo(t: T1) {
  t.a = undefined
}

foo(t1) // function foo changes prop "a" back to undefined
// t1.a = undefined  ; this should generate an error

console.log(t1.a.b) // ... "a" is still considered defined here and TS compiles successfully :/
// This will result in a runtime error.

I would anticipate TypeScript to acknowledge that foo manipulates the mutable value t1 (properties in T1 do not have the readonly flag). foo could potentially alter the value (which it does in this case), so the compiler should reset all previous narrowing types of t1 and report a in console.log(t1.a.b) as possibly undefined once again.

My inquiry now is: How intricate is this control flow mechanism and what rules does it adhere to in the above example? You can also view the code in action through this link. Thank you!

Answer №1

The function foo is a typical side-effectful function that accepts an argument, modifies it, and returns void. TypeScript follows a top-down flow in the code, checking if foo receives the correct argument but not controlling how it affects the scope above. The only way to communicate the changes made by foo to the outer scope is by returning the modified value. Currently, the only information available to the outer scope about foo is that it requires a T1 type argument without returning anything.

It's worth mentioning that this implicit argument modification approach is considered an anti-pattern because foo is an unsafe function to use without providing any information at the type level due its lack of return value. By returning the modified value, we can inform the higher scope about the changes. See the example below:

let t1: T1 = {} // change to let
t1.a = { b: "foo" }

function foo(t: T1) {
    t.a = undefined
    return t; // returning the modified value
}

t1 = foo(t1) // using the mutated value returned from the function
console.log(t1.a.b) // will result in error as 'a' is undefined

This approach allows TypeScript to understand the return value of foo and its effects on the outer scope.


A better practice would be to make foo a pure function (though this is just an opinion):

function foo(t: T1) {
    return {...t, a: undefined};
}

Answer №2

It is important to verify each time whether a possesses the property b:

let result = ('b' in object.a) ? object.a.b : undefined;

For more information on the in operator, check out this link.

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