2. 19&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, 6.15

3. 19&20-3 Pointer 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”.

4. 19&20-4 We 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.

5. 19&20-5 There are two C operators that are necessary when using pointer variables. & - the address operator returns the address of a variable x 1000 3 ptrX 1000 ptrX 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; 1004

6. 19&20-6 Example: int x; int * pt; pt = &x; /* another way to assign an address to a pointer variable */ Suppose the address of the variable x is 9640 Then, the effect of pt = &x; will be: xpt Address 9640

7. 19&20-7 int 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. x pt Address 9640 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:

8. 19&20-8 double *ptr; ptr = NULL; Assigning a NULL value (zero) to a pointer A null pointer points to nothing. We often depict it as ptr /*called a null pointer*/

9. 19&20-9 1. Problem Definition Write 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 integers Output= no value is returned to the calling function, but the values of the called variables should be swapped. 3. Develop Algorithm Pass 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.

10. 19&20-10 /* C Function to swap values. */ #include 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; }

11. 19&20-11 x 1000 Address Main Memory before call to swap 1 y 2 Main Memory while executing swap, after temp = *ptrX; but before *ptrX = *ptrY ; 1000 Address 1 ptrX 1000 3000 2 x y 1004 3008 3004 1004 ptrY temp 1

12. 19&20-12 x y Main Memory while executing swap, after *ptrX = *ptrY ; but before *ptrY = temp; 1000 Address 2 ptrX 1000 3000 2 1004 3008 3004 1004 ptrY temp 1

13. 19&20-13 x y Main Memory while executing swap, after *ptrY = temp; but before return; 1000 Address 2 ptrX 1000 3000 1 1004 3008 3004 1004 ptrY temp 1

14. 19&20-14 1. 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.

15. #include 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)

16. /* 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 */

17. 19&20-17 unixprompt>./a.out Please enter the number of array elements: 5 1 2 3 4 5 The average is:3.000000 -2.000000 0.000000 1.000000 2.000000 unixprompt>

18. 19&20-18 The 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. 19&20-19 In C, we can dynamically allocate storage with either of the following two functions (both are in ). 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 operations.

20. 19&20-20 In both cases above (calloc or malloc) return memory locations to the system ("memory manager") with: free (ptr); Example: Declare ptr as a pointer variable to data- type double and use dynamic memory allocation (calloc or malloc) to allocate 100 elements of data- type double. double * ptr; ptr = calloc(100,sizeof(double)); or ptr = malloc(100*sizeof(double));

21. 19&20-21 A string literal (e.g. “A – your course grade\n”) is actually represented in C by a pointer to an address in memory that holds the first byte of the array of characters. Therefore the following declaration of the pointer variable “string” is valid: char * string = "A – your course grade\n"; and to print the value of “string”, use the “%s” conversion specifier, printf("%s",string); Although a pointer can be used as an array name, we cannot modify the values of “string” by using the square brackets of array notation. string[0] = ‘B’; /* error !!! */ The error is due to the fact that the literal “A – your course grade\n” is stored in a location in memory that cannot be accessed by pointers.

22. 19&20-22 One use of char * pointers is ragged arrays. In lecture 15-36 we used the following code fragment to declare and initialize the array “morse” as : typedef char string[5]; string morse[26] = {".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..","--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.." }; An alternative would be the use of a ragged “morse” array declared as: typedef char * string; string morse[26] = {".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..","--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.." };

23. 19&20-23 The later declaration of the array “morse” is called ragged since each element of the array, morse[0], morse[1], … takes up exactly the number of bytes necessary in memory to hold the initialized strings whereas in the previous declaration of “morse” each element of the array takes exactly five character values in memory no matter what length of the initializing string. As discussed on slide 21, we cannot change the values of the ragged “morse” array. But in the “Morse Code” problem we know that the morse strings should never be changed.