6 Replies to “Array”

  1. Would this really be considered a copy, if the array is just referencing the old array? Changes made on the new array will be reflected on the old array.

    START OF TEST CODE

    // Setting up the Original Array called old_array

    var old_array:Array = [];

    old_array[0][0] = “0x0”;
    old_array[0][1] = “0x1”;

    old_array[1][0] = “1×0”;
    old_array[1][1] = “1×1”;

    old_array[2][0] = “2×0”;
    old_array[2][1] = “2×1”;

    // 'Copying' old_array array into the newly created new_array

    var new_array:Array = old_array.concat();

    trace(“Old Array:”);
    trace(old_array.concat());

    trace(“New Array:”);
    trace(new_array.concat());

    trace(” “)
    trace(“Removing a Node from New Array”);
    trace(” “)

    // Removing an element from new_array

    new_array[2].splice(1,1);

    trace(“Old Array:”);
    trace(old_array.concat());

    trace(“New Array:”);
    trace(new_array.concat());

    END OF TEST CODE

    You maybe able to discern from the Output data that the new_array does not independently exist, but instead is coupled with old_array. Any changes made to new_array will be reflected on old_array.

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  3. Yeah you are right if the array of numbers is streaming from some source.But actually we have got a fixed array of numbers. we are going to divide this array into 2 arrays of least difference in their sums.To put in other words we are going to divide this array into 2 arrays such that the sum of each array is close to the (sum of the original array /2).So the algorithm would be o(n) instead of o(nlogn) as in your case.
    t-total sum of original array
    s-0
    for each element in array
    {
    if((s+element)

  4. Is it still an LED matrix if each LED gets its own driver? Also the parallax coming from the big gap between the touch panel and the array looks like it could be problematic. The software looks pretty slick tho.

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