previous / next within a range of numbers

hey all.. i need a function that would return a previous and next numbers, but only within my numbers range. so, for example, if my range is from 0 to 7, and im on 6 - next should return 7. if im on 7 - next should return 0 (it circled back to it).

same for previous, if im on 0, previous should be 7. I think modulo can be used to figure this out, but cant figure out how. the function should take 3 arguments- current number we are on, maximum number and if we 开发者_如何学Pythonare going back or forward. something like

getPreviousOrNext(0, 7, "next" or "prev" )

thanks!!!

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Use modulo..

function getPreviousOrNext(now, max, direction) {
    totalOptions = max + 1; //inlcuding 0!

    newNumber = now; // If direction is unclear, the number will remain unchanged
    if (direction == "next") newNumber = now + 1;
    if (direction == "prev") newNumber = now + totalOptions - 1; //One back is same as totalOptions minus one forward

    return newNumber % totalOptions;
}

(could be shorter, but this makes it more understandable)

Edit: The "now + totalOptions - 1" prevents us from going into negative numbers (-1 % 7 = -1)

Edit2: Ouch, there was a small error in the code..."If direction is unclear, the number will remain unchanged" was not correct!

Edit3: And for a bonus, this is how I would have written it before reading Code Complete ;-) (assumes it's 'next' whenever it's not 'prev'). This is ugly and beautiful in one :

function getPreviousOrNext(now, max, direction) {
   return (now + ((direction=="prev")?max:1)) % (max + 1);
}

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is this a homework assignment?
i wouldn't use modulo, a handful of if/ternary statements should be sufficient.

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var cycle_range = function (high, current) {
    return new function () {
       this.next = function () {
           return current = (current+1) % (high+1);
       };

       this.previous = function () {
           return current = (current+high) % (high+1);
       };
    }
};

cycle_range(7, 0).next() // 1

var the_range = cycle_range(7, 0);
the_range.next() // 1
the_range.next() // 2
the_range.previous() //1
the_range.previous() //0
the_range.previous() //7