Type wrongly assumed to be undefined | number in this case #31603

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1. AnyhowStep commented on May 26, 2019

   

   AnyhowStep

   on May 26, 2019

   Last edited by AnyhowStep

   Contributor

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   Probably a design limitation because it requires narrowing the type of an expression (instead of a variable, or property), but just wanted to point out the obvious; you can restructure your code to work around it.

   let a: number | undefined;
   let b: number | undefined;
   const tmp = a || b;
   
   if (tmp) {
   	const c = tmp;
   }

   Although, it would be cool to have narrowing for side-effect free expressions. Not sure if there's much benefit to it.

   I never use logical operators with non-boolean types, though.
2. fatcerberus commented on May 26, 2019

   

   Bruce Pascoe (fatcerberus)

   on May 26, 2019

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   Yeah, the problem is that a || b, from the compiler's point of view, doesn't prove anything individually about a or b, so it can't actually narrow them. At the time the assignment to c happens the types are still as follows:

   a :: number | undefined
   b :: number | undefined

   So as far as the compiler is concerned, both could be undefined at the time the assignment to c happens. We know that's not the case since a || b requires at least one of them to be non-falsy, but the type narrowing mechanism isn't quite that nuanced. It doesn't track dependencies between variables--the same reason this doesn't work:

   let x: number | undefined
   let isNumber = typeof x === 'number';
   if (isNumber) {
       // x :: number | undefined
       x += 1;  // error
   }

   👍React with 👍7Reacted by S. B. Tam, poseidonCore, Joe Calzaretta, Titian Cernicova-Dragomir, Lisiur Day, Kamran Ayub and Livio Brunner
3. Ryan Cavanaugh (RyanCavanaugh)

   added

   Design LimitationConstraints of the existing architecture prevent this from being fixedConstraints of the existing architecture prevent this from being fixed

   on May 28, 2019
4. kamranayub commented on Dec 20, 2019

   

   Kamran Ayub (kamranayub)

   on Dec 20, 2019

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   I believe we just encountered this limitation and were scratching our heads:

   Playground link

   interface TestInterface {
     value?: string;
   }
   
   function foo({ value }: TestInterface) {
     const isValidValue = !!value && /bar/.test(value);
     const thing = isValidValue ? value.split('x') : [];
                               // ^ Object is possibly 'undefined'
   }

   This is a simplified example of what we were trying within our code. "Clearly" if value is falsy, there's no way it would be undefined when checking isValidValue.
5. Ryan Cavanaugh (RyanCavanaugh)

   closed this as completedon Jun 1, 2021