This document discusses pointers in C programming. It defines pointers as variables that contain memory addresses and can be used to indirectly access and manipulate other variables. It covers declaring and initializing pointer variables, pointer operators like address-of and dereference, using pointers to call functions, and arrays of pointers. Examples are provided to illustrate pointers to integers, characters, and how pointers can traverse arrays by incrementing the pointer. Exercises at the end demonstrate assigning a pointer variable the address of another variable and printing the value of the object pointed to.

1. Chapter 6 – POINTERS Pointer Pointers Declaration & Initialization Pointer Operators Calling Function using Pointers Array of Pointers Array of string

2. Objectives To have acquired the ability to design and manipulate pointer variables. To be able to design and use user-defined data types- structures

3. 6.1 Pointers A variable has a name, an address, a type and a value Pointers are language constructs that allow programmers to directly manipulate the address of variables. Simulate call-by-reference Close relationship with arrays and strings

4. 6.1.1 Declaration & Initialization Pointer variable Contain memory addresses as their values Normal variables contain a specific value (direct reference) Pointers contain address of a variable that has a specific value (indirect reference) Indirection – referencing a pointer value

5. 6.1.1 Declaration & Initialization(Cont..) Pointer declarations * used with pointer variables int * myPtr; Declares a pointer to an int (pointer of type int * ) Multiple pointers require using a * before each variable declaration int * myPtr1, * myPtr2; Can declare pointers to any data type Initialize pointers to 0 , NULL , or an address 0 or NULL – points to nothing ( NULL preferred)

6. 6.1.1 Declaration & Initialization (Cont..) Pointer Initializations 2 ways :- i) Assigning address of other variable ii) assigning value of other variable directly Method (i) Example : int val = 30; int *value =&val; Method (ii) Example : int *value ; *value = 30;

7. 6.1 2 Pointer Operators

8. 6.1.2 Pointer Operators (Cont..)

9. 6.1.2 Pointer Operators (Cont..) Short Notes : i) int *px; px = pointer to an integer ii) int x; x is an integer iii) px = &x; px gets the address of x or px point to x iv) y = *px; y gets the content of whatever px points to

10. Example 1 int age, *ag; …(i) age = 30; ….(ii) ag =&age; ….(iii) Situation in Computer Memory . After (i) age 1002 ag 1004 B) After (ii) age 1002 ag 1004 ? ? 30 ? C) After (iiI) age 1002 ag 1004 30 1002 Assume that the following codes are added after (iii). Predict the output values ? printf(“%d\n”, age); printf(“%d\n”, ag); printf(“%d\n”, *ag);

11. Example 2

12. Example (Cont..) Refer to Add Note : Ex. 1 & 2

13. 6.1.3 Calling Function using Pointers Refer to Add Note : Ex. 3

14. 6.1.4 Arrays of Pointers

15. Example char *name[5] = { { “Ahmad”}, {“Maria”}, {“Halim”}, {“Nora”}, {“Lina”} }; To print 1 st String printf(“%s”, *name); To printf 2 nd String printf(“%s”, (*name+1));

16. Example Program /* Example : Arrays of pointers*/ /*Increments a pointer through an integer array*/ #include<stdio.h> main( ) { int number[] = { 10,20,30,40,50}; int *p = number; /* The pointers points to the start of the array*/ printf(“%d \n”, *p); p++; printf(“%d \n”, *p); p++; printf(“%d \n”, *p); p++; printf(“%d \n”, *p); p++; printf(“%d \n”, *p); }

17. Exercise Write a C statement to accomplish the following tasks. Assume that the floating point variable number1 has been declared. Define the variable fPtr to be a pointer to an object of type float. Assign the address of variable number1 to pointer variable fPtr. Print the value of the object pointed to by fPtr.

18. Exercise Write a C statement to accomplish the following tasks. Assume that the floating point variable number1 has been declared. Define the variable fPtr to be a pointer to an object of type float. Assign the address of variable number1 to pointer variable fPtr. Print the value of the object pointed to by fPtr. Answers : float *fPtr; fptr = &number1; printf(“The value of object pointed is %d”, *fPtr);