Removing null check using if statement in typescript

Question

Having the following code:

function A(arg1?: string, arg2?: number) {
  if (!arg1 && !arg2) {
    throw new Error('At least one of two parameters must be present');
  }
  B(arg1 || arg2);
}

function B(arg3: string | number) {
  // does something. we don't really care
}

Typescript throws the following compilation error for the expression B(arg1 || arg2):

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

Type 'undefined' is not assignable to type 'string | number'. ts(2345)

However, in function A I make sure that at least one of the arguments is not undefined, throwing an error in that case. That means that in the expression B(arg1 || arg2) the argument will never be undefined, it will always be either number or string.

Is something wrong in my assumption? Is there any way of making typescript understand it?

You could try B(arg1 || arg2 || 0) Just to ensure the compiler has a fallback, that never is used. — Nikolaus
– Nikolaus, Commented Apr 29, 2019 at 13:13

Aron · Accepted Answer · 2019-04-29 13:27:08Z

7

If you were doing a truthy check on a single value TS would indeed correctly infer that the value is no longer undefined

if (arg1) {
  arg1; // arg1: string
}

When you check two variables though, TS doesn't know for certain that either of them individually are truthy, so arg1 and arg2 are still string | undefined and number | undefined respectively.

One option is to assign arg1 || arg2 to a variable before the if statement so the truthy inference works correctly.

function A(arg1?: string, arg2?: number) {
    const arg = arg1 || arg2; // arg: string | number | undefined
    if (arg) {
        // arg: string | number
        B(arg);
    } else {
        throw new Error('At least one of two parameters must be present');
    }
}

function B(arg3: string | number) {
    // does something. we don't really care
}

answered Apr 29, 2019 at 13:27

Aron

9,3187 gold badges37 silver badges60 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

hoangdv · Accepted Answer · 2019-04-29 13:24:20Z

2

A new ! post-fix expression operator may be used to assert that its operand is non-null and non-undefined in contexts where the type checker is unable to conclude that fact.

What's new

If you sure that at least one of args is not undefined, you could use ! syntax like this:

function A(arg1?: string, arg2?: number) {
  if (!arg1 && !arg2) {
    throw new Error('At least one of two parameters must be present');
  }
  B(arg1! || arg2!); // In compile time `arg1` and `arg2` are not `undefined`, in runtime maybe, but you sure the case  never come true :)
}

function B(arg3: string | number) {
  // does something. we don't really care
}

answered Apr 29, 2019 at 13:24

hoangdv

16.3k4 gold badges31 silver badges54 bronze badges

Collectives™ on Stack Overflow

Removing null check using if statement in typescript

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related