How to check with is_page if 404.php is loaded?

Question

I want to check if current page match at least one of two conditions. I don't want to show js code into about page and into 404 page. For the rest of my pages i show the js code. So, I use ((!is_page(element1) AND !is_page(element2)) then show the js code. Element1 is a slug of one specific page and element2 is 404.php page. I don't know the slug or name of 404 page to indicate in if statement.

Answer 1

I think this would do :

if ( is_page( 'element1.php' ) || is_404() ) {
// do that  if page is using template page element1.php or is 404
}

Answer 2

A 404 page probably doesn't have a slug. It is created automagically like the tag and category pages archive pages

Now, is_page doesn't tell you anything about the template being used, which you seem to be thinking about checking for when you say and "and element2 is 404.php page". is_page_template will. So...

if (is_page('element1') || is_page_template('404.php')) {
  // stuff
}

Will trigger if page-slug == element1 is loaded or if a page is loading that uses 404.php. But 404.php is not a necessary theme file so looking for the template may not be what you want. You probably want to use is_404 as in Simon's answer (which you should "accept" if you go with is_404).