6
\$\begingroup\$

This function adds a number of hours to a 24 hour clock:

/**
 * @param {Integer} now The current hours
 * @param {Integer} add The number of hours to add
 */
function addHours(now, add){
    var h = (now + add) % 24;

    return h < 0 ? 24 + h : h;
};
  • addHours(2, 5) //7
  • addHours(23, 5) //4
  • addHours(23, -5) //18
  • addHours(72, 3) //3
  • addHours(0, 0) //0

The output should always be an integer in the range of 0 - 23.

I can't see any problem with it, but maybe you can. Please let me know if it can be improved or if there are any bugs.

\$\endgroup\$
2
  • \$\begingroup\$ Other that the inappropriate parameter name now, there is not much you could say about it code-wise. However what I'd like to question its usefulness, which seems very limited. \$\endgroup\$
    – RoToRa
    Commented Jul 3, 2014 at 13:02
  • \$\begingroup\$ @RoToRa I'm using as part of a set of unit tests. I have to test some other functions that manipulate time, and this one helps keep the tests readable. \$\endgroup\$
    – Drahcir
    Commented Jul 3, 2014 at 13:17

1 Answer 1

5
\$\begingroup\$

Tiny question ;)

  • As @Rotora mentioned, now is not a fantastic parameter name, it conveys that I need to only pass who late it is 'now', add is a verb, also not brilliant as parameter naming goes

  • addHours( -500 , -100 ) returns -0, is that what you want ?

  • You have 4 lines of comment, 3 lines of code, 1 blank line, perhaps you have to much comment ? Try to not need comments by working harder on parameter names.

  • Since adding is commutative ( order does not matter ), I would consider naming the function sumHours and name the parameters hours1 and hours2

    function sumHours(hours1, hours2){
        var sum = (hours1 + hours2) % 24;
        return sum < 0 ? 24 + sum : +sum;
    };
    
\$\endgroup\$
2
  • \$\begingroup\$ Thanks for finding the -0 bug, actually I didn't know -0 was even a number. Also your other points are helpful. \$\endgroup\$
    – Drahcir
    Commented Jul 3, 2014 at 13:29
  • \$\begingroup\$ Negative zero might not be that big of a problem for you, as it behaves very much like positive zero. You would have to go out of your way to test 1 / -0 === Number.NEGATIVE_INFINITY to detect negative zero. \$\endgroup\$ Commented Jul 5, 2014 at 1:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.