I suggest using an array with as many elements as the longest row you need. To keep it simple, keep two copies of the array, and calculate each element of the "new" array as the sum of the corresponding element, plus the previous element, of the "old" array. Then copy the information back for the next step.
This type of sorting can b performd by simply transferring all the matrix elements in a single dimension array of 1X16 size and then sorting this array and then transferring the elements back to 4X4 matrix. You can also treat the 4x4 matrix as a simple array using pointers and, thus, not need to transfer from matrix to array and back. Example, using ellipses (...) to simulate indentation for clarity... int matrix[4][4] = {...some values...} int *element; int flag = 1; while (flag == 1) { /* simple bubble sort */ ... flag = 0; ... /* loop from first element to next to last element */ ... for (element = &matrix[0][0]; element < &matrix[3][3]; element ++) { ... ... if (*element > *(element + 1)) { ... ... ... flag = 1; ... ... ... *element ^= *(element + 1); /* exclusive or swap */ ... ... ... *(element + 1) ^= *element; ... ... ... *element ^= *(element + 1); ... ... } ... } }
A single dimensional array is an array of items. A two-dimensional array is an array of arrays of items.
An irregular dimensional array is a special type of multi-dimensional array.First we must understand that a multi-dimensional array is just an array of arrays. Each element in the array is, itself, an array of elements.A regular multi-dimensional array will be an array of size n, with each element containing a separate array of size m. That is, each sub-array has the same size.An irregular multi-dimensional array will be a multi-dimensional array in which each sub-array does not contain the same number of elements.Regular array:array[0] = new array{0, 1, 2}array[1] = new array{3, 4, 5}array[2] = new array{6, 7, 8}array[3] = new array{9, 10, 11}This regular array is an array of size 4 in which each sub-array is of size 3.Irregular array:array[0] = new array{0, 1, 2}array[1] = new array{3, 4}array[2] = new array{5, 6, 7}array[3] = new array{8, 9, 10, 11}This irregular array is an array of size 4 in which the size of each sub-array is not the same.
Option 1) Use a temporary variable: int x = array[i]; array[i] = array[i+1]; array[i+1] = x; Option 2) Use bit operators: array[i] ^= array[i+1] ^= array[i];
you use lines and arrows and draw through numbers
like this
it is 3 squared
25
15 is the answer The + will be used as the "dots" for the array Array:
The answer will depend on what the problem is: some can be solved using an array but for others, arrays are a complete waste of time.
it provide array taste and tantalizing aromas that draw you into the kitchen
you have to pick if you want to draw a 1*5 or a 5*1. then you have eaither put from up to down way or a across way thankyou
I suggest using an array with as many elements as the longest row you need. To keep it simple, keep two copies of the array, and calculate each element of the "new" array as the sum of the corresponding element, plus the previous element, of the "old" array. Then copy the information back for the next step.
9 sets of five pennies
One example is making 7 cans of juice in each of six rows: ••••••• ••••••• ••••••• ••••••• ••••••• •••••••
This type of sorting can b performd by simply transferring all the matrix elements in a single dimension array of 1X16 size and then sorting this array and then transferring the elements back to 4X4 matrix. You can also treat the 4x4 matrix as a simple array using pointers and, thus, not need to transfer from matrix to array and back. Example, using ellipses (...) to simulate indentation for clarity... int matrix[4][4] = {...some values...} int *element; int flag = 1; while (flag == 1) { /* simple bubble sort */ ... flag = 0; ... /* loop from first element to next to last element */ ... for (element = &matrix[0][0]; element < &matrix[3][3]; element ++) { ... ... if (*element > *(element + 1)) { ... ... ... flag = 1; ... ... ... *element ^= *(element + 1); /* exclusive or swap */ ... ... ... *(element + 1) ^= *element; ... ... ... *element ^= *(element + 1); ... ... } ... } }