Introduction to Pointers
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Understanding Pointers: Why, Declaration, Assignment, Access, and Uses, Slides of Computer Engineering and Programming
An introduction to pointers in c programming language. It covers the concept of pointer variables, their declaration, assignment, access, and two uses of the pointer operator. It also includes examples and explanations of the & and * operators.
Typology: Slides
2012/2013
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Introduction to Pointers
Overview
1. Why Pointer Variables?
2. Boxes Again
3. What is a Pointer Variable?
4. Declaring a Pointer Variable
5. Assignment to Pointers
6. Accessing a Pointer
7. Two Uses of *
8. More Examples
9. Coding Call by Reference
- Boxes Again
The box occurs at memory address 1232
(say):
int x; x
x
3. What is a Pointer Variable?
A pointer variable can
contain the memory address
of another variable.
- Assignment to Pointers
The operator for returning the address of a
variable is &
int x = 5; int *countptr; countptr = &x;
Assigment in Pictures int x = 5; x
1232 (say)
int *countptr; countptr
countptr = &x; countptr
x
countptr
Usually, drawn as:
Initialising a Pointer
int *xptr;
float *yptr; xptr = NULL; yptr = NULL;
- Accessing a Pointer int c = 13; int *countptr; countptr = &c; printf("The value of countptr is %lu\n", countptr);
Result:
- Two Uses of * int *countptr;
means countptr can point to an integer variable
All other occurrences of * mean dereference:
printf("...", *countptr);
- More Examples
a gptr
a gptr
a gptr
int a = 3; *gptr; gptr = &a; *gptr = 7;
The & and * Operators Fig. 7. #include <stdio.h> int main() { int a = 7 ; int aptr; aptr = &a; / continued on next slide */
printf("Address of a is %lu\n", &a); printf("Value of aptr is %lu\n\n", aptr); printf("Value of a is %d\n", a); printf("Value of aptr is %d\n\n", aptr); printf("%s\n", "Proving that * and & are complements of each other."); printf("&aptr = %lu\n", &aptr); printf("*&aptr = %lu\n", *&aptr); return 0; }
Declarations and Initialisations int i = 3, j = 5, *p = &i, *q = &j, *r; double x; Expression Equivalent Expression Value p == &i p == (&i) 1 (true) p = i + 7 p = (i + 7) illegal **&p *p (^3) r = &x r = (&x) illegal
- Coding Call by Reference