Ifology - how to write this statement better?

Question

I'm wondering how to write the if statement in the following block in a better way. It's supposed to operate when $a is 14, 22, 30 and for all following values at intervals of 8, up to some limit. The current way is obviously not good since the action must be performed each time that periodic pattern is fulfilled, and this would require many, many OR operators.

if($a == 14 || $a == 22 || $a == 30 || $a == 38 || $a == 46... until some number){
    do something...
}

Generally you should try if (isSomething($a)) { do something ... } and then you can solve that problem in isSomething, ideally with the solution @Winston Ewert proposed. — back2dos, Sep 25 '14 at 06:49

score 15 · Accepted Answer · answered Sep 24 '14 at 21:01

15

if($a % 8 == 6 && $a >= 14 && a <= some_number)

In general, if you want periodicity, think modulus.

answered Sep 24 '14 at 21:01

Winston Ewert

24,862

score 2 · Answer 2 · answered Sep 25 '14 at 02:21

If your test values are not in any specific sequence (are not powers or two, not all even, not all odd, etc) you can use a lookup table to avoid the long if chains:

int test_values[] = { 14, 22, 30, 38, 46, ... };

for (i = 0; i < test_values.length; i++)
{
    if (input == test_values[i])
    {
        do something...
    }
}

This way it is also very easy to add more values by just expanding the array. The array of cases can also be initialized from an external config file.

score 1 · Answer 3 · answered Sep 25 '14 at 01:09

1

In Python this is fairly descriptive:

if a in range(14, some_number, 8):
    do something

answered Sep 25 '14 at 01:09

Matthew Finlay

493
2
8

Ifology - how to write this statement better?

3 Answers3