PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 2319&20-2•Know how to declare pointer variables. •Understand the & (address) and *(indirection) operators.•Dynamic Memory Allocation•Related Chapter: ABC 6.2 – 6.6, 6.8, 6.10, 6.11, 6.1519&20-3Pointer variables are variables that store memory addresses.Pointers variables are useful in passing storage addresses in function calls (call-by-reference) and for applications involving dynamic data structures (e.g., linked lists) Example: int *intPtr; float *floatPtr;declares intPtr to be a pointer variable to an object of type integer. and floatPtr to be a pointer variable to an object of type float. The data type for intPtr is, int * ,read as “pointer to an int”. The data type for floatPtr is, float * ,read as “pointer to a float”.19&20-4We can write the pointer variable declaration as int* intPtr; or as int * intPtr;Note that when we declare more than two pointer variables in one line each pointer name requires an asterisk: int *ptr1, *ptr2;otherwise we are just declaring a regular variable , not a pointer variable.19&20-5There are two C operators that are necessary when using pointer variables.& - the address operator returns the address of a variablex 10003ptrX1000ptrX holds the address of x.We say ptrX “points” to x. We will apply the & operator only to variables, e.g. &77 or &(x+1) are not valid.The declaration of ptrX initializes the variable ptrX = &x;Example:int x = 3;int * ptrX = &x;100419&20-6Example:int x;int * pt;pt = &x; /* another way to assign an address to a pointer variable */ Suppose the address of the variable x is 9640Then, the effect of pt = &x; will be:x ptAddress9640964019&20-7int x;int * pt;pt = &x;To assign the value “3” to the variable x in C: x = 3;We can use the “ * ” operator to indirectly reference x. We can assign “3” to x by:*pt = 3;Here we use the fact that “pt” points to integer values.xptAddress96409640 3* - means "a pointer to" and is called an indirection operator or dereferencing operator, since a pointer "indirectly" references a value in a storage location. Example:19&20-8double *ptr;ptr = NULL;Assigning a NULL value (zero) to a pointerA null pointer points to nothing. We often depict it asptr/*called a null pointer*/19&20-91. Problem DefinitionWrite a function “swap” that has two input integer arguments. This function should swap the values of the variables. 2. Refine, Generalize, Decompose the problem definition(i.e., identify sub-problems, I/O, etc.)Input = two integersOutput= no value is returned to the calling function, but the values of the called variables should be swapped.3. Develop AlgorithmPass the addresses of the variables and not their values. In the swap function the parameters that hold the addresses are pointers. Use the indirection operator to swap the values.19&20-10/* C Function to swap values. */#include <stdio.h>void swap(int * , int *); void main(void) /* function header */{ int x = 1; int y = 2; swap(&x,&y); /* pass the addresses of x and y */ printf("x = %i , y = %i \n",x,y); /* prints x = 2 , y = 1 */}void swap(int *ptrX , int *ptrY) /* pointer variables */{ int temp = *ptrX; *ptrX = *ptrY ; *ptrY = temp;}19&20-11x 1000AddressMain Memory before call to swap 1y2Main Memory while executing swap, after temp = *ptrX; but before *ptrX = *ptrY ; 1000Address1ptrX100030002xy10041004300830041004ptrYtemp119&20-12xyMain Memory while executing swap, after *ptrX = *ptrY ; but before *ptrY = temp; 1000Address2ptrX1000300021004300830041004ptrYtemp119&20-13xyMain Memory while executing swap, after *ptrY = temp;but before return; 1000Address2ptrX1000300011004300830041004ptrYtemp119&20-141. Problem Definition Write a program that reads a list of real numbers into an array from the keyboard (or a file using Unix redirection). The array can be of arbitrary length but the user must first specify the length of the array. The program then calculates the average value, and then prints a list of differences. The differences are computed by taking the original values in the list minus the average. 2. Refine, Generalize, Decompose the problem definition (i.e., identify sub-problems, I/O, etc.) Input = Since the length of the array is specified at run-time we must use Dynamic Memory Allocation. This is accomplished in C by the use of the built-in function calloc (or malloc). Use data-type double to hold the values. Output= The average and the list of differences.#include <stdio.h>#include <stdlib.h>void main(void){ int k,num; double * datValptr; /* pointer used in DMA */ double datAve; double sum = 0.0; /* prompt the user for the number of array elements */ printf("Please enter the number of array elements:"); scanf("%i",&num); /* use calloc to allocate a block of memory */ datValptr = calloc(num,sizeof(double)); /* verify that calloc was successful */ if (datValptr = = NULL) {printf("Memory not allocated!\n");return; }(continued on next slide)/* read the values and compute the average */ k = 0; for(k=0;k<num;++k) { scanf("%lf", &datValptr[k]);/* use pointer as array name */ sum += datValptr[k]; } /* compute the average */ datAve = sum /num; printf("The average is:%f \n", datAve); /* compute and print the diff list */ for(k=0;k<num;++k) { printf("%f\n", datValptr[k]-datAve); } /* end of for*/ /* free up any memory that was dynamically allocated */ free(datValptr);} /* end of main */19&20-17unixprompt> ./a.outPlease enter the number of array elements: 51 2 3 4 5The average is:3.000000-2.000000-1.0000000.0000001.0000002.000000unixprompt>19&20-18The array name in C is assigned the address of the first element of the array. The following code will assign the address of x[0] to xPtr :float x[50];float * xPtr; xPtr = x;The assignment statement above is equivalent to:xPtr = &x[0];19&20-19In C, we can dynamically allocate storage with either of the following two functions (both are in <stdlib.h>).• malloc(numberOfBytes)• calloc(numberOfItems, itemSize)Both functions return a pointer to the address of the block of storage that has been allocated. If no memory is available, a NULL value will be returned.Function calloc returns a contiguous block of locations that are initialized to 0 and that can be referenced as array locations, using pointer