typescript checking using type guards

Question

I have a method which launches a modalWindow which takes an element as input (using $event.target) or a number. I am trying to perform a check on the type of the input but the check always treats element as a number even if it's not. Whether or not element is a number the console.log is outputted and the else never runs.

launchModal(element: number | any) {

    if (<number>element) {

      console.log('ele as num', element);
    } else {
      const elementId = (event.target as Element).id;
      const elementIdArray = String(elementId).split('-');
      this.currentRow = parseInt(elementIdArray[2], 10);

    }
    this.loadEquipmentFullRowDataRequest(this.currentRow);

  }

I following the documentation for advanced types at the official site

Motti · Accepted Answer · 2017-09-12 11:27:34Z

3

In typescript the type system is compile time only, in runtime the code is transpiled to JavaScript and all the type information is lost.

When you say if (<number>element) the cast doesn't actually do anything, it just allows you to write code that refers to element as a number (here you aren't doing anything to element so it's meaningless).

In order to check if the value you got is really a number you have to use JavaScript methods:

if(typeof(element) === 'number')

answered Sep 12, 2017 at 11:27

Motti

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Shane · Accepted Answer · 2017-09-12 11:28:51Z

1

if(<number>element) does not type check if the element is a number, once compiled it will just check if element is not null or undefined.

Try the following:

const launchModal = (elem: number | HTMLElement) => {
    if (!elem) {
        return;
    }

    let row = elem as number;
    if (elem instanceof HTMLElement) {
        row = parseInt(elem.id.split('-')[2], 10);
    }

    // load row data
}

e.g. :

launchModal(10);

launchModal(document.getElementById('...'));

answered Sep 12, 2017 at 11:28

Shane

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