(array.length - 1) will find the index of the last element in an array (or -1 if the array is empty).
To dynamically allocate an array, use 'void *realloc(void *array, size_t size)', where array is the pointer you plan on reallocating and size is the new length of the array in bytes.One-dimensional array:{int len = 10;int *array = NULL;array = (int*)realloc(array, len * sizeof(int));}Two-dimensional array:{int xlen = 10;int ylen = 10;int **array2d = NULL;array2d = (int**)realloc(array2d, xlen * sizeof(int*));/* After reallocating array2d, do the same to its contents. */int index;for (index = 0; index < xlen; index++) {array2d[index] = (int*)realloc(array2d[index], ylen * sizeof(int));}array2d[5][5] = 24;/* Clean up. */for (index = 0; index < xlen; index++) {free(array2d[index]);}free(array2d);return 0;}
Basically, &array[i]; That is, the memory location for an array object with index i. Or, you can do: (array + i);
Array subscripts always have a zero-based index. In languages that allow an n-based index, the index is simply offset by n elements, so the compiler subtracts n from the given index to obtain the zero-based index. Arrays are always zero-based because the first element of an array is found zero elements from the start of the array.
using multidimensional array
By design; it makes the compiler's work easier. 1-based array's addressing-function: Address (array, index) = Address (array) + (index-1)*Elemsize(array) 0-based array's addressing-function: Address (array, index) = Address (array) + index*Elemsize (array)
A matrix IS an array so it is impossible to multiply a matrix without array. The answer to the multiplication of two matrices need not be an array. If the first matrix is a 1xn (row) matrix and the second is an nx1 (column) matrix, then their multiple is a 1x1 matrix which can be considered a scalar.
(array.length - 1) will find the index of the last element in an array (or -1 if the array is empty).
A key is the name of a variable in an array ($array["key"]) and the index is the position it's at ($array = ["key" => 0], the index would be 0). Keys and indices are the same if the array is not associative though ($array = [true], the key holding the value true is named 0 and is at index 0).
matrix
To dynamically allocate an array, use 'void *realloc(void *array, size_t size)', where array is the pointer you plan on reallocating and size is the new length of the array in bytes.One-dimensional array:{int len = 10;int *array = NULL;array = (int*)realloc(array, len * sizeof(int));}Two-dimensional array:{int xlen = 10;int ylen = 10;int **array2d = NULL;array2d = (int**)realloc(array2d, xlen * sizeof(int*));/* After reallocating array2d, do the same to its contents. */int index;for (index = 0; index < xlen; index++) {array2d[index] = (int*)realloc(array2d[index], ylen * sizeof(int));}array2d[5][5] = 24;/* Clean up. */for (index = 0; index < xlen; index++) {free(array2d[index]);}free(array2d);return 0;}
a matrix
Basically, &array[i]; That is, the memory location for an array object with index i. Or, you can do: (array + i);
Array subscripts always have a zero-based index. In languages that allow an n-based index, the index is simply offset by n elements, so the compiler subtracts n from the given index to obtain the zero-based index. Arrays are always zero-based because the first element of an array is found zero elements from the start of the array.
using multidimensional array
A matrix in mathematics is a rectangular array of quantities or expressions set out by rows and columns.
Nothing, but a two dimensional array can be used to represent a matrix.