Detect months with 31 days

Question

Is there an analogous form of the following code:

if(month == 4,6,9,11)
{
  do something;
}

Or must it be:

if(month == 4 ||         


        

Answer 1

If you're using C or Java, you can do this:

switch (month) {
  case 4:
  case 6:
  case 9:
  case 11:
    do something;
    break;
}

In some languages, you could even write case 4,6,9,11:. Other possibilities would be to create an array [4,6,9,11], some functional languages should allow something like if month in [4,6,9,11] do something;

As Lior said, it depends on the language.

EDIT: By the way, you could also do this (just for fun, bad code because not readable):

if ((abs(month-5) == 1) || (abs(month-10) == 1)) do_something;

Answer 2

For dates I use Joda Time mentioned earlier, but I understand if it's not applicable for you. If you just want it to look nice, you can first define a list with values that you're interested in and then check if your month is in that list:

// This should be a field in a class
// Make it immutable
public static final List<Integer> LONGEST_MONTHS = 
        Collections.immutableList(Arrays.asList(4,6,9,11));

// Somewhere in your code
if(LONGEST_MONTHS.contains(month)) {
    doSomething();
}

Answer 3

Simpe Aux function (C#) that gives you the number of days of a given month:

    private int GetDaysInMonth(DateTime date)
    {
        return new DateTime(date.Year, date.Month, 1).AddMonths(1).AddDays(-1).Day;
    }

Hope it helps

Answer 4

A complete solution, also taking leap year into account.

private static int getDaysInMonth(LocalDateTime localDateTime) {

    int daysInMonth = 0;
    int year = localDateTime.getYear();
    int month = localDateTime.getMonth().getValue();

    switch (month) {
        case 1: case 3: case 5:
        case 7: case 8: case 10:
        case 12:
            daysInMonth = 31;
            break;
        case 4: case 6:
        case 9:case 11:
            daysInMonth = 30;
            break;
        case 2:
            if(((year % 4 == 0) && !(year % 100 == 0) || (year % 400 == 0))) {
                daysInMonth = 29;
            } else {
                daysInMonth = 28;
            }
            break;
        default: System.out.println("Invalid month");
            break;
    }

    return daysInMonth;
}

Answer 5

A rather literal translation into Java would be:

if (Arrays.binarySearch(new int[] { 4, 6, 9, 11 }, month) >= 0) {

I don't know what is so special about 4, 6, 9 and 11. You are probably better off using an enum, together with EnumSet or perhaps a method on the enum. OTOH, perhaps JodaTime does something useful.

Answer 6

In Icon, you can do

if month = (4|6|9|11) then ...

You can also do

if a < b < c then ...

See also

• Wikipedia: Icon (programming language)