CODEWITHSUNDEEP
typescript
user19069181
user19069181

Reputation:

Object possibly null despite checking

Why is TS complaining

Object possibly null

despite my check?

let sum = (data: [...(number | null)[]]): number => {
  let result: number = 0;
  for (let i = 0; i < data.length; i++)
    if(data[i]!==null){
      result += data[i]; // here is complain Object possibly null
    }
    
  return result;
};

console.log(sum([1, 2, null, 4]));

Upvotes: 1

Views: 1243

Answers (2)

Lars Flieger
Lars Flieger

Reputation: 2562

WARNING: If the iterable object has a similar getter, then the first check in the if condition will call it once and then the assignment will call it a second time. The value can change between the if and the assignment. e.g.: Math.random() > 0.5 ? null : 42

You could use a Non-null assertion:

result += data[i]!;

Since you can 100% say in this context this is not null.

Typescript Playground

Upvotes: 0

brunnerh
brunnerh

Reputation: 185445

Since this is accessing a property of an object via key variable, TS does not recognize that both statements refer to the same value. You can store the value in a temporary constant to fix this:

const item = data[i];
if (item !== null) {
  result += item;
}

But I would not recommend using for loops anyway. Either use for ... of or a functional approach using filter and reduce.

for (const item of data) {
  if (item !== null) {
    result += item;
  }
}

Examples of various types of access that rightly or wrongly may cause errors:

const o = {
  x: 12 as number | null,
}
let key = 'x' as keyof typeof o;

if (o.x != null) {
  console.log(o.x + 1); // OK
}

if (o.x != null) {
  doSomething();
  console.log(o.x + 1); // Wrong but allowed
}

if (o[key] != null) {
  console.log(o[key] + 1); // Error
}

function doSomething() {
  o.x = null;
}

const arr: (number | null)[] = [12];

if (arr[0] != null) {
  console.log(arr[0] + 1); // OK
}

if (arr[0] != null) {
  doSomething2();
  console.log(arr[0] + 1); // Wrong but allowed
}

let i = 0;
if (arr[i] != null) {
  console.log(arr[i] + 1); // Error
}


function doSomething2() {
  arr[0] = null;
}

Playground

Upvotes: 5

Related Questions