1 structures, dynamic memory allocation. 2 agenda structures definition & usage pointers to...

Structures, Structures, Dynamic Dynamic Memory Memory

Allocation Allocation

Agenda

Structures Definition & usage Pointers to structures Arrays and pointers in structures

Dynamic Memory Allocation

Structures

‘Logical entities’ Examples: complex numbers, dates,

student records, geometric objects, etc’

Each is composed of a number of variables

Structures

struct: collection of variables, gathered into one variable

Defines new data types Memory Variables in a struct are called

members or fields

Example – complex numbers

Definition of a new ‘type’ that represents a complex number:

struct complex { int real; int img; };

Once we define a structure, we can use it as any type:

struct complex num1, num2, num3;

Access structure members If A is of some structure with a member

named x, then A.x is that member of Astruct complex C;C.real = 0;

If A is a pointer to a structure with a member x, then A->x is that member of the variable pointed by A (same as (*A).x)struct complex *pc = &C;pc->real = 1;

Convenient usage with typedef

typedef struct complex_t { int real; int img; } complex;

A new variable type: “complex” Saves writing “struct complex” every

time! Usage: complex num1, num2;

Examples (AddComplex.c)

complex AddComp (complex x, complex y) { complex z;

z.real = x.real + y.real; z.img = x.img + y.img;

return z;}

Structures are passed to functions “by value”

A copy of the structure is passed

AddComplex – step by step

complex a, b, c;

printf(“…");scanf("%lf%lf",&(a.real),&(a.img));printf(“…");scanf("%lf%lf",&(b.real),&(b.img));

c = AddComp(a,b);

printf(“result = %g+%gi\n",c.real,c.img);return 0;

… …

complex a, b, c;

c = AddComp(a,b);

… …

complex a, b, c;

c = AddComp(a,b);

1.0 2.0

… …

complex a, b, c;

c = AddComp(a,b);

1.0 2.0

… …

complex a, b, c;

c = AddComp(a,b);

1.0 2.0

3.0 4.0

… …

complex a, b, c;

c = AddComp(a,b);

1.0 2.0

3.0 4.0

… …

complex AddComp(complex x, complex y){ complex z;

return z;}

1.0 2.0

3.0 4.0

… …

return z;}

1.0 2.0

3.0 4.0

… 6.0

return z;}

1.0 2.0

3.0 4.0

4.0 6.0

return z;}

1.0 2.0

3.0 4.0

4.0 6.0

complex a, b, c;

c = AddComp(a,b);

1.0 2.0

3.0 4.0

4.0 6.0

complex a, b, c;

c = AddComp(a,b);

1.0 2.0

3.0 4.0

4.0 6.0

Exercise Implement the MultComplex

function – Input - two complex numbers Output – their multiplication Definition: x=a+ib and y=c+id then:

z = xy = (ac-bd)+i(ad+bc)

Write a program that uses the above function to multiply two complex numbers given by the user

Solution (MultiplyComplex.c)

complex MultiplyComp(complex a, complex b) {complex c;

c.real = a.real*b.real - a.img*b.img; c.img = a.real*b.img + a.img*b.real;

return c;}

More on Structures

Structures cannot be compared using the == operator They must be compared member by member Usually this will be done in a separate

function Structures can be copied using the =

operator Member-wise copy

Example (Is_In_Circle.c)

int IsInCircle(dot *p_dot, circle *p_circle) { double x_dist,y_dist;

x_dist = p_dot->x - p_circle->center.x; y_dist = p_dot->y - p_circle->center.y;

if((x_dist * x_dist + y_dist * y_dist) <= (p_circle->radius * p_circle->radius))

return 1;

return 0;}

Is_in_circle – step by step

printf(“Enter dot\n");scanf("%lf%lf",&d.x,&d.y);printf("Enter circle center\n");scanf("%lf%lf",&c.center.x,&c.center.y);printf("Enter circle radius\n");scanf("%lf",&c.radius);

if (IsInCircle(&d, &c))printf("dot is in circle\n");

elseprintf("dot is out of circle\n");

… …

d (dot)

c (circle)

… …

center (dot) radiu

… …

d (dot)

c (circle)

… …

center (dot) radiu

1.0 2.0

d (dot)

c (circle)

… …

center (dot) radiu

1.0 2.0

d (dot)

c (circle)

… …

center (dot) radiu

1.0 2.0

d (dot)

c (circle)

0.0 0.0

center (dot) radiu

1.0 2.0

d (dot)

c (circle)

0.0 0.0

center (dot) radiu

1.0 2.0

d (dot)

c (circle)

0.0 0.0

center (dot) radiu

1.0 2.0

d (dot)

c (circle)

0.0 0.0

center (dot) radiu

Is_in_circle – step by stepint IsInCircle(dot *p_dot, circle *p_circle){ double x_dist,y_dist;

x_dist = p_dot->x - p_circle->center.x; y_dist = p_dot->y - p_circle->center.y; if (x_dist*x_dist + y_dist*y_dist <= p_circle->radius*p_circle-

>radius) return 1;

return 0;}

1.0 2.0

(circle)

0.0 0.0

center (dot) radiu

x_dist

y_dist… …

p_circle

p_dot1024756

>radius) return 1;

return 0;}

1.0 2.0

(circle)

0.0 0.0

center (dot) radiu

x_dist

y_dist1.0 …

p_circle

p_dot1024756

>radius) return 1;

return 0;}

1.0 2.0

(circle)

0.0 0.0

center (dot) radiu

x_dist

y_dist1.0 2.0

p_circle

p_dot1024756

>radius) return 1;

return 0;}

1.0 2.0

(circle)

0.0 0.0

center (dot) radiu

x_dist

y_dist1.0 2.0

p_circle

p_dot1024756

>radius) return 1;

return 0;}

1.0 2.0

(circle)

0.0 0.0

center (dot) radiu

x_dist

y_dist1.0 2.0

p_circle

p_dot1024756

1.0 2.0

d (dot)

c (circle)

0.0 0.0

center (dot) radiu

Exercise

Write a struct that represents a date (day, month, year)

Write a function that increments the datevoid IncDate(Date *d);

For example – 31.12.05 -> 1.1.06

Solution

IncDate.c

Structures containing arrays A structure member that is an array

does not ‘behave’ like an ordinary array ‘=‘: the array is copied element by

element It is not the address that gets copied! For example - array_member.c

Reminder – ordinary arrays can’t be copied simply by using the ‘=‘ operator They must be copied using a loop

Structures containing arrays Same behavior when passing the

structure to a function Changing the array inside the function won’t

change it in the calling function Reminder – when passing an ordinary

array to a function, all that gets passed is the address of its first element Hence every change to the array within the

function, changes the array in the calling function

Pointers are another matter

If the member is a pointer all that gets copied is the pointer (the address) itself For example, pointer_member.c

Make sure that you understand what you do!

Dynamic Memory Allocation

Dynamic Memory Allocation: Motivation

Array variables have fixed size (e.g.: int – 4 bytes, char – 1 byte)

This can’t be changed after compilation

It is not always known how many elements we will need in runtime

We would like to be able to dynamically allocate memory

The malloc function

void *malloc(unsigned int n);

The function malloc is used to dynamically allocate n bytes

malloc returns a pointer to the allocated area on success, NULL on failure

The malloc function

void *malloc(unsigned int n);

We should always check whether memory was successfully allocated

Remember to #include <stdlib.h> Allocated memory must be freed

(later)

Usage Example

int n, *p;printf("How many numbers do you want to

enter?\n");scanf("%d",&n);

p = (int *)malloc(n*sizeof(int));

if(p == NULL) {printf("Memory allocation failed!\n");return 1;

free(p);

Why casting?

The casting in p=(int *) malloc(n*sizeof (int));

is needed because malloc returns void * :void *malloc(unsigned int nbytes);

The type void * specifies a general pointer, which can be cast to any pointer type.

What is this ‘sizeof’ ?

The sizeof operator gets a variable or a type as an input and outputs its size in bytes:

double x; s1=sizeof(x); /* s1 is 8 */ s2=sizeof(int) /* s2 is 4 */

Free the allocated memory segment

void free(void *ptr);

We use free(p) to free the allocated memory pointed to by p

If p doesn’t point to an area allocated by malloc, a run-time error occurs

Free the allocated memory segment

void free(void *ptr);

Always remember to free the allocated memory once you don’t need it

Otherwise, you may run out of memory – a common bug that is hard to detect

Example

dynamic_reverse_array.c

Example (dynamic_reverse_array.c)

int i, n, *p;printf("How many numbers do you want to enter?\n");scanf("%d",&n);

p = (int *)malloc(n*sizeof(int));if(p == NULL) {

printf("Memory allocation failed!\n");return 1;

}printf("Please enter numbers now:\n");for(i=0; i<n; i++)

scanf("%d", &p[i]);

printf("The numbers in reverse order are - \n");for(i=n-1; i>=0; i--) printf("%d ",p[i]);free(p);

Allocating memory within a function

Dynamic allocation of memory is not deleted when we leave the scope / exit a function

This is why we need to free it Now we are able to allocate

memory within a function and use it outside

Example (another_strcpy.c)

char *another_strcpy(char *src) {char *dst; int len, i;

len=strlen(src);dst=(char*)malloc(sizeof(char)*(len+1));if(dst == NULL) {

printf("Memory allocation failed!\n");

return NULL;}

for(i=0;i<=len;i++)dst[i] = src[i];

return dst;}

Exercise (@ home) Implement the function my_strcat –

Input – two strings, s1 and s2 Output – a pointer to a dynamically allocated

concatenation (‘shirshur’) For example: The concatenation of “hello_”

and “world!” is the string “hello_world!” Write a program that accepts two strings

from the user and prints their concatenation Assume input strings are no longer than a

100 chars

Solution

my_strcat.c (my_strcat2.c)

What’s wrong with this?char *my_strcat(char *str1, char *str2){

int len;char result[500]; /* Let’s assume this is large enough */

len = strlen(str1);

strcpy(result, str1);strcpy(result+len, str2);

return result;}

Exiting the program

void exit(int status); Sometimes an error occurs and we want

the program to immediately exit The exit function closes all open files,

frees all allocated memory, and exits the program

Equivalent to calling ‘return’ within main Remember to #include <stdlib.h> See strcpy_with_exit.c

1 structures, dynamic memory allocation. 2 agenda structures definition & usage pointers to...

complex c c

realimg b realimg c

real y

c complex addcomp complex

complex y

complex z z

realimg x

realimg y realimg z

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