# Ifology - how to write this statement better? [closed]

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...
}
``````
• Are the numbers in the sequence all of the form 8k+6? – Ryan Sep 25 '14 at 3:16
• 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 6:49

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

In general, if you want periodicity, think modulus.

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.

In Python this is fairly descriptive:

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