I have a simple check where I want to check if the given variable is >=0.

public print(value: any): void {
    if(value >= 0) {
      console.log('Greater than zero')
    }
  }

The catch here is when the ining variable has value null, then it will bee truthy and log the statement. Is there a clean way to avoid it, but not adding extra checks?

I have a simple check where I want to check if the given variable is >=0.

public print(value: any): void {
    if(value >= 0) {
      console.log('Greater than zero')
    }
  }

The catch here is when the ining variable has value null, then it will bee truthy and log the statement. Is there a clean way to avoid it, but not adding extra checks?

• javascript
• typescript
• logical-operators

• 9 What's wrong with an extra check? if(value !== null && value >= 0) { – ibrahim mahrir Commented Jun 20, 2020 at 19:36
• 2 Number(null) evaluates to 0 so any slick tricks won't apply – Józef Podlecki Commented Jun 20, 2020 at 19:39
• 1 There is if(value?.valueOf() >= 0) if you want your coworkers to throw stationery in your direction. – adiga Commented Jun 20, 2020 at 19:42
• 2 How about if (typeof value === 'number' && value >= 0) for the check? – Alberto Rivera Commented Jun 20, 2020 at 19:44
• 1 just use parseInt check if not NaN and don't bother – Józef Podlecki Commented Jun 20, 2020 at 19:52

4 Answers 4

You can employ a type guard that will assure the piler that you're not handling a null but a number. Moreover, it will make the code more correct, since with value: any this means you might get a boolean or a string passed in:

public print(value: any): void {
  if (typeof value === "number") {
    //value is definitely a number and not null
    if (value >= 0) {
      console.log('Greater than zero')
    }
  }
}

Playground Link

Now the code specifically verifies that you do get a number and then checks if it's more than or equal to zero. This means that a null or a non-number value would not be processed.

The type guard condition can be bined with the other for brevity:

public print(value: any): void {
  if (typeof value === "number" && value >= 0) {
    console.log('Greater than zero')
  }
}

Playground Link

Or extracted on its own to just reduce the nesting:

public print(value: any): void {
  if (typeof value !== "number")
    return;

  //value is definitely a number and not null
  if (value >= 0) {
    console.log('Greater than zero')
  }
}

Playground Link

I don't see why you don't want to add a null-check.

An alternative is to use number instead of any but it will work only if your ts.conf enables strict null checks.

function print(value: number): void {
    if(value >= 0) {
      console.log('Greater than zero')
    }
}

print(null) // won't pile with strict null checks

If you codebase does not allow the use of null, just use undefined and use an implicit conversion, like so:

public print(value: any): void {
    if(value != undefined && value >= 0) {
        console.log('Greater than zero')
    }
}

This works because null == undefined (the double equals creates a type conversion, while the triple equals does not).

In JavaScript, I usually use the following:

`${value}` >= 0

// or

parseInt(value) >= 0

in TypeScript you can most likely use:

public print(value: any): void {
  if (+`${value}` >= 0) {
    console.log('Not less than zero')
  }
}