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Chapter 6 – POINTERS

• Pointer• Pointers Declaration & Initialization• Pointer Operators• Calling Function using Pointers• Array of Pointers

• Array of string

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Objectives

• To have acquired the ability to design and manipulate pointer variables.

• To be able to design and use user-defined data types- structures

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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

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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

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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)

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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;

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6.1 2 Pointer Operators

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6.1.2 Pointer Operators (Cont..)

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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

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Example 1

int age, *ag; …(i)age = 30; ….(ii)ag =&age; ….(iii)

Situation in Computer Memory.A) 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);

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Example 2

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Example (Cont..)

Refer to Add Note : Ex. 1 & 2

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6.1.3 Calling Function using Pointers

Refer to Add Note : Ex. 3

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6.1.4 Arrays of Pointers

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Example

char *name[5] = { { “Ahmad”},

{“Maria”},

{“Halim”},

{“Nora”},

{“Lina”} };

To print 1st String

printf(“%s”, *name);

To printf 2nd String

printf(“%s”, (*name+1));

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Example Program

1. /* Example : Arrays of pointers*/2. /*Increments a pointer through an integer array*/3. #include<stdio.h>4. main( ) {5. int number[] = { 10,20,30,40,50};6. int *p = number; /* The pointers points to the start of the

array*/7.8. printf(“%d \n”, *p);9. p++;10. printf(“%d \n”, *p);11. p++;12. printf(“%d \n”, *p);13. p++;14. printf(“%d \n”, *p);15. p++;16. printf(“%d \n”, *p);17. }

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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.

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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);