TypeScript null checks not paying attention to array.length check

Question

When compiling TypeScript using strict null checks, the following does not type check even though it's fine:

const arr: number[] = [1, 2, 3]
const f = (n: number) => { }
while (arr.length) {
    f(arr.pop())
}

The compilation error is:

Argument of type 'number | undefined' is not assignable to parameter of type 'number'. Type 'undefined' is not assignable to type 'number'.

It seems that the compiler isn't smart enough to know that arr.pop() will definitely return a number.

Some questions:

Why isn't the compiler smarter? Would adding smarter null-checking for this kind of case be incredibly difficult, or is it something straightforward that the TS team hasn't implemented yet?
What's the most idiomatic way to write the above that still type-checks?

Re 2, the best I can come up with is to add a superfluous check to the body of the loop:

while (arr.length) {
    const num = arr.pop()
    if (num) { // make the compiler happy
        f(num)
    }
}

It is correct, when you do a pop and it does not have any elements to return. It returns a null, that is why it has this as a return type "number | undefined" in the case if the element is empty. — ngeksyo
– ngeksyo, Commented Apr 5, 2018 at 2:50
I tried build your code on [email protected] with "awesome-typescript-loader": "^3.2.3", it doesn't have any compiling error. — FisNaN
– FisNaN, Commented Apr 5, 2018 at 2:59

artem · Accepted Answer · 2018-04-05 02:53:42Z

3

Yes adding this kind of smartness to compiler is considered difficult, see this comment on the very exact issue that describes your problem.

In the meanwhile, you can use non-null assertion - postfix ! - to tell the compiler that you know that the value is not null:

const arr: number[] = [1, 2, 3]
const f = (n: number) => { }
while (arr.length) {
    f(arr.pop()!)
}

answered Apr 5, 2018 at 2:53

artem

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