If you mean how do you create an array with 16 elements, there are two ways:
int a[16]; /* fixed size array of 16 integer elements */
int* b = malloc(16*sizeof(int)); /* variable length array with (initially) 16 integer elements */
Remember that variable length arrays allocated on the heap must be released as soon as they are no longer required:
free (b);
b=NULL;
the length of the array
length
You can make arrays with any number of dimensions (depending on RAM limitations, of course). However, internally, a two-dimensional array (for example) is stored as an array of arrays; that is, each first-level array contains an array of the second level. Similarly with higher dimensions.
Since an array cannot contain a negative number of items, the size of an array must be at least 0. So if you ever tried to retrieve the element at a negative index in an array, it would automatically be understood to be out-of-bounds.
Full representation of an array begins from the index 0 and ends at n-1 where n is the number of variables of the array.
You can create an array with some elements, get the factors of a number (such as 16), and while you get each of the factors, place the factors in the array. Every time you find a factor, divide the number by this factor before you search the next factor (to avoid getting the same factor over and over again).
13
2D array of size 2x8 and 1D array of size 16
The number of arrays you can make with the number 16 depends on how you define "arrays." If you're referring to the factors of 16, they are 1, 2, 4, 8, and 16, which can form rectangular arrays of various dimensions (e.g., 1x16, 2x8, 4x4). In terms of combinations or arrangements of the number 16 in an array (like in permutations), the possibilities would be significantly greater, depending on the context and constraints you apply.
13...13x1.
A square array is an array in which the number of rows is the same as the number of columns.
In order to determine which of four number is bigger you need 16 if statements arranged in a nested if statement. That exceeds the complexity of just sorting it in an array. Here is a solution using a simple array sort. int array[] = {7, 2, 27, 4}; int i, swap; do { swap = 0; for (i = 0; i < 4; ++i) { if (array[i]<array[i+1]) { swap=array[i]; array[i]=array[i+1]; array[i+1]=swap; swap=1; } } } while (swap == 1); printf ("%d\n", array[0]);
Ah, honey, in C, you can get the number of elements in an array by dividing the total size of the array by the size of one element. So, if you have an array of integers, you can do something like int size = sizeof(array) / sizeof(array[0]); and voilà, you've got the number of elements. Just be careful with those pesky pointers and make sure you're not trying to count elements in a pointer instead of an actual array.
The minimum number of swaps required to sort an array is equal to the number of inversions in the array.
the length of the array
length
It is an array with the same number of rows and columns.