02 Arrays And Memory Mapping

Linear Data Structures Arrays Strings Structures Stacks Queues Link List (Logically Linear)‏

Representation of Single Dimensional Arrays Dimension ? Whole array is stored in a single contiguous memory block. Address of the first element is called the base address and it is also the start address of array. The address of the ith element is given by the following formula: Address i = base_address + i * size_of_element Arrays are very efficient way of organizing data since accessing array elements requires O(1). elem n-1 … elem i … elem 2 elem 1 elem 0

float sum (float *list, int n)‏ { float temp = 0; int i; for (i = 0; i < n; i++)‏ temp = temp + *(list + i); return temp; } float sum (float list[], int n)‏ { float temp = 0; int i; for (i = 0; i < n; i++)‏ temp = temp + list[i]); return temp; }

Representation of Two Dimensional Array Can be called Table Needs two dimensions - One for Rows - Other for columns Two representations Row Major Column Major

Representations of Table Row Major ( First subscript changes least frequently and last subscript changes most frequently)‏ (i,j)th element = Base Address + (i * N + j) * size of array Column Major ( Last subscript changes least frequently and first subscript changes more frequently)‏ (i,j)th element = Base Address + (j * M + i) * size of array

void add (int a[][], int b[][], int c[][], int rows, int cols)‏ { int i, j; for (i=0; i<rows; i++)‏ for (j=0; j<cols; j++)‏ c[i][j] = a[i][j] + b[i][j]; } What’s wrong?

const int N = 100; void add (int a[][N], int b[][N], int c[][N], int rows, int cols)‏ { int i, j; for (i=0; i<rows; i++)‏ for (j=0; j<cols; j++)‏ c[i][j] = a[i][j] + b[i][j]; } Why?

Two Dimensional Array Offset of a[i][j]? 54 345 106 0 99 82 76 22 64 38 2 3 6 3 5 1 User’s view (abstraction)‏ 54 345 106 0 99 82 76 22 64 38 2 3 6 3 5 1 System’s view (implementation)‏

Two Dimensional Array Offset of a[i][j]? Number of elements in each row? i th row total elements in the previous rows? j th element in i th row Offset (in units) N i * N i * N + j The addressing formula needs to know the number of columns but does not need the number of rows! 54 345 106 0 99 82 76 22 64 38 2 3 6 3 5 1 User’s view (abstraction)‏

Row major – C, Pascal, C++, etc. Column Major – FORTRAN Ada - indefinite Two Dimensional Array

Row Major Slice along the 1 st dimension to get an array of N-1 dimensions Continue until you are left with one dimension only. Offset of a[d 0 ] [d 1 ] …[d n-1 ] Multi-Dimensional Arrays A[D 0 ] [D 1 ]…[D n-1 ] (…((d 0 *D 1 + d 1 )*D 2 + d 2 )*D 3 + …) + d n-1

Simple View General formula in Row Major A[s1,s2,s3,...sn] = Base Address + ((s 1 * D 2 *D 3 *..D n ) + (s 2 * D 3 *D 4 *..D n ) + (s n-1 *D n ) + S n ) * Size of array General formula in Column Major A[s1,s2,s3,...sn] = Base Address + ((s n * D 1 *D 2 *..D n-1 ) + (s n-1 * D 1 *D 2 *..D n-2 ) + (s 2 *D 1 ) + S 1 ) * Size of array

Symmetric Diagonal Triangular Sparse * make programs of all these properties Special Matrices

Ordered Lists One of the most common data object is an ordered list. An ordered list has elements with definite ordering. The simplest example of an ordered list is an ordered pair which has only two elements. Examples of ordered lists: Months of the years (Jan, Feb, Mar, Apr, May,…., Dec)‏ English Alphabets (A, B, C, …, Z)‏ Words in a sentence (“This is a book on data structures.”)‏ Names of the students in a class stored in the order of their roll numbers.

Operations Defined on an Ordered List Find the length of list. Check if the list is empty. Traverse the list in some order. Get the i th elements in the list. Change the value of the i th element. Insert a new element at the i th position. Delete the i th element. *Make programs of all these properties

Representation of an Ordered List The easiest way to represent an ordered list is by an one-dimensional array where we store the i th element of the list in the array element with index i. This is called sequential or linear mapping – mapping a one-dimensional vector onto a one-dimensional space.

class LinearList { public: LinearList(int s); // constructor ~ LinearList () {delete [ ] ListArray; } // destructor bool add (int value); bool remove (int index); bool change(int index, int value); bool get(int index, int & value); void printAll(); bool isFull() {return size == maxSize ;} bool isEmpty() {return size == 0; } int length() {return size;} private: int maxSize; // max List size int *ListArray; int size; // no. of elements in the List void readjust(int index); void moveForward(int index); void moveBackward(int index); }; Linear List Using Arrays

LinearList :: LinearList(int s)‏ { maxSize = s; ListArray = new int[maxSize]; size = 0; } void LinearList :: printALL()‏ { for (i = 0; i < size; i++)‏ cout << ListArray[i] << newline; }

bool LinearList :: get(int index, int & value) { if (index < size && index >= 0) { value = ListArray[index]; return true; } else return false; }

bool LinearList::add(int value)‏ { if (! isFull() ) { for (int i = size; i >= 1; i--) { if (ListArray[i-1] > value)‏ ListArray[i] = ListArray[i -1]; else break; } ListArray[i] = value; size++; return true; } else return false; }

bool LinearList::remove(int index)‏ { if (index < size && index >= 0) { for (int i = index; i < size - 1; i++)‏ ListArray[i] = ListArray[i +1]; size--; return true; } else return false; }

bool LinearList :: change(int index, int value) { if (index < size && index >= 0) { ListArray[index] = value; readjust(index); return true; } else return false; } void LinearList :: readjust(int index) { if ((index < (size - 1)) && (ListArray[index] > ListArray[index + 1]))‏ moveForward(index); else if ((index > 0) && (ListArray[index] < ListArray[index - 1]))‏ moveBackward(index); }

void LinearList :: moveForward(int index) { int temp = ListArray[index]; int i = index; while ((i < (size - 1)) && (ListArray[i] > ListArray[i + 1])) { ListArray[i] = ListArray[i+1]; i = i + 1; } if (i != index) ListArray[i] = temp; } void LinearList :: moveBackward(int index) { int temp = ListArray[index]; int i = index; while ((i > 0 ) && (ListArray[i] < ListArray[i - 1])) { ListArray[i] = ListArray[i-1]; i = i - 1; } if (i != index) ListArray[i] = temp; }

Assignment# 1 last date of submission : 02-09-2008 (Tuesday)‏ Obtain an addressing formula for elements a[i,j] in the upper triangular matrix of order nxn. You can assume that this upper triangular is stored row by row Arrays - Read a matrix , find upper triangular matrix, find diagonal matrix, find determinant of a matrix Strings - Read a string , find the length of string, a particular word, replace a particular word with new word. - Read an array of names of students and sort it. Structures - Make an array of structures (Roll-No, Name, Marks, Grade) , Display all records, find all the students with particular grade , find the record of a particular student, delete record of a particular student

02 Arrays And Memory Mapping

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