undefined not treated as falsy in typescript-CodePudding

const f = (x: number) => { }

const a : (number | undefined)[] = [1,undefined,3];

for (let i = 0; i < a.length; i   ) {
    if (a[i]) {
        f(a[i]);
    }
}

In the example above, a is an array that can hold an element which can be either number of undefined.

In the loop, I want to only access the element if it is not undefined. My understanding is that if (a[i]) would filter every undefined values (also 0 too, but let's not think of that here).

So I thought the typescript would think the type of a[i] inside the if would be simply number. But if I get this error when calling f.

Argument of type 'number | undefined' is not assignable to parameter of type 'number'.
  Type 'undefined' is not assignable to type 'number'.
'x' is declared but its value is never read.

Why is the type of a[i] cannot be deduced to number with the if check above?

CodePudding user response：

The below code will work fine, it is because typescript does not deduce what type a[i] would result in, in the if condition itself. Creating a variable and then checking the same would work. This is a known issue as you can access both arrays and object with the square bracket [].

const f = (x: number) => { }

const a : (number | undefined)[] = [1,undefined,3];

for (let i = 0; i < a.length; i   ) {
    let j = a[i];
    if (j) {
        f(j);
    }
}