Programming Structures and Dynamic Allocation. Dynamic Memory Allocation The memory requirement of...

Programming

Structures andDynamic Allocation

Dynamic Memory Allocation

The memory requirement of our program is not always known in advanceArrays have fixed size determined at compile

time We would like to allocate at runtime

The Memory

Two Parts:1. Stack

Memory is allocated when entering a function Deallocation when the function terminates

2. Heap Programmer control allocation and deallocation

Memory Management

stdlib.h Two operations for memory management

1. Allocation void* malloc(…);

2. Deallocation void free(…);

The malloc Function

void* malloc(unsigned int nBytes);

malloc is used to dynamically allocate nBytes in memory

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

You should always check whether memory was successfully allocated

Remember to #include <stdlib.h>

The free Function

void free(void *ptr);

Use free(p) to deallocate memory pointed by p

If p doesn’t point to an area allocated by malloc an error occursNo partial deallocation

Always remember to free the allocated memory once you are done with it

Exampleint main(void){ int i, n, *p;

printf("How many numbers will you enter?\n"); scanf("%d", &n);

/* Allocate an int array of the proper size */ p = (int*)malloc(n * sizeof(int)); if (p == NULL) { printf("Memory allocation failed!\n"); return 1; }

/* Free the allocated space */ free(p);

return 0;}

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 ‘sizeof’ ?

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

sizeof always calculates the size of the type double x;s1 = sizeof(x); /* s1 is 8 */s2 = sizeof(int) /* s2 is 4 */

Example

char name[25];char* address = (char*)malloc(25);int i;

printf("sizeof(name) = %d\n", sizeof(name));

printf("sizeof(address) = %d\n", sizeof(address));

if (sizeof(&i) == sizeof(address))

printf("True\n");

printf("False\n");

Returning Allocated Memory

Since allocated memory is not deallocated upon function termination we can return its address.

Duplicate a stringchar* strdup(const char* str){ char* dup = (char*)malloc((strlen(str)+1) * sizeof(char)); if (dup != NULL) { strcpy(dup, str); } return dup;}

Exercise

Implement the functionchar* my_strcat(const char *s1, const char *s2); Output – a pointer to a dynamically allocated

concatenation of s1 and s2 For example: The concatenation of “hello_” and

“world!” is the string “hello_world!”

Test your function

Solutionchar* my_strcat(const char *s1, const char *s2){ int len; char *result = NULL;

len = strlen(s1) + strlen(s2) + 1;

result = (char*)malloc(len * sizeof(char)); if (result == NULL) { printf(“Memory allocation failed!\n"); return NULL; }

strcpy(result, s1); strcat(result, s2);

return result;}

What’s Wrong Here?

char* my_strcat(const char *s1, const char *s2){ int len; char result[500]; /* assume this is enough */

strcpy(result, s1); strcpy(result + strlen(s1), s2);

return result;}

Pitfalls

Accessing un-initialized pointer

int* p;*p = 3;

Using de-allocated memory

int* p = (int*)malloc(SIZE * sizeof(int));... /* do something with p*/free(p);*p = 0;

Structures

Not much in the real world comes to us as simple data types. (e.g. Car, Book, Bank account)

We want to be able to manipulate logical entities as a whole

The individual pieces of the data may exist in one of the basic forms

How do you keep track of what pieces of the data belong together?

Examples

BookTitle, Authors, ISBN, Publisher

Bank AccountOwners, Balance, Credit

CarMake, Model, Year

A convenient way of grouping several pieces of related information together.

A collection of variables under a single name.

Structures - User Defined Types

Structures - Example

struct point{ double x; double y;};

struct rectangle{ struct point low_left; struct point up_right;};

struct student{ char id[10]; char* name; int avg_grade;};

struct point{ double x, y;};

Defining a struct

struct struct-name{

type field-name1; type field-name2; type field-name3;

Example - Complex Numbers

Structure declaration

Variable definition Define variables of type complex

struct complex { double real, img;};

struct complex c1, c2;

Field Access

Use the . (dot) operator to access a field in a structure

If comp is of type complex then to access the real part we write

comp.realand to access the imaginary part we write

comp.img

Example

Printing a complex number

printf("%g + %gi", comp.real, comp.img);

Input from the user

scanf("%lf", &comp.real);

Typdef

Introduce synonyms for typestypedef int Boolean;typedef char* String;

Boolean b;String str = "Text";

Typdef and Structs

Instead of writing struct complex everywhere we can create a shorter synonym using typedef. typdef struct complex Complex

Then use it Complex c1, c2;

We can combine the typdef with the structure declaration:

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

This is the common way to define new types

Typedef Even Shorter

Structures and Functions

Function can receive structures as their parameters and return structures as their return value

Parameter passing is done by value copy

Similar to basic types

Structures and Functions - Example

Complex make_complex(double real, double img){ Complex temp;

temp.real = real; temp.img = img;

return temp;}

Complex add_complex(Complex c1, Complex c2){ return make_comlex(c1.real + c2.real, c1.img + c2.img);}

add_complex – step by step

int main(void){ Complex a, b, sum;

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

sum = add_complex(a, b);

printf("result = %g+%gi\n", sum.real, sum.img);

return 0;}

… …

return 0;}

… …

return 0;}

1.0 2.0

… …

return 0;}

1.0 2.0

… …

return 0;}

1.0 2.0

3.2 4.5

… …

return 0;}

1.0 2.0

3.2 4.5

… …

Complex add_complex(Complex x, Complex y){ Complex sum;

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

return sum;}

1.0 2.0

3.2 4.5

… …

return sum;}

1.0 2.0

3.2 4.5

… 6.5

return sum;}

1.0 2.0

3.2 4.5

4.2 6.5

return sum;}

1.0 2.0

3.2 4.5

4.2 6.5

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

return 0;}

1.0 2.0

3.2 4.5

4.2 6.5

return 0;}

1.0 2.0

3.2 4.5

4.2 6.5

return 0;}

1.0 2.0

3.2 4.5

4.2 6.5

Assignment

Structures can be assigned (copied) using the assignment operator “=” Bitwise copy – copying the content of one

structure’s memory onto another c1 = add_complex(…);

1001001110101010010101010101110101110101

0000000000000000000000000000000000000000

1001001110101010010101010101110101110101

Arrays in Structs

The entire array is part of the structure When passing the struct to a function (by

value) Changing the array field won’t change the

original array

Pointers in Structs

When copying a struct containing a pointer only the pointer is copied (shallow copy)

Not what the pointer points to we should take extra care when

manipulating structures that contain pointers

Comparison

Structures cannot be compared using the equality operator “==” They must be compared member by

member Usually this will be done in a separate

function

is_equal_complex(Complex c1, Complex c2){ return (c1.real == c2.real) && (c1.img == c2.img);}

Exercise

Implement a multiply_complex function:

Complex multiply_complex(Complex x, Complex y);

Note: if x = a + bi and y = c + di then: z = xy = (ac - bd) + (ad + bc)i

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

Solution

Complex make_complex(double real, double img){ Complex temp; temp.real = real; temp.img = img;

return temp;}

Complex multiply_complex(Complex a, Complex b){ return make_complex(a.real * b.real - a.img * b.img, a.real * b.img + a.img * b.real);}

Pointers to Structures

Same as with other types Pointer definition

structure-name * variable-name; Complex* comptr;

Complex c, *cptr;cptr = &c;

defines a complex and a pointer to a complex

assign the address of the complex variable to the complex pointer

To access the fields we can write:

(*pcomp).real(*pcomp).img

Complex comp;Complex *pcomp = &comp;

Accessing Fields Using Pointers

Alternative Syntax

Pointer to structures are extremely common

Alternative notation as a shorthand p→member-of-structure → is minus sign followed by >

Examplepcomp→realpcomp→img

Example

typdef struct point { double x, y;} Point;

typdef struct rect { Point pt1, pt2;} Rect;

Rect r, *rp = &r;

r.pt1.x(*rp).pt1.xrp->pt1.x

equivalent

Example

typedef struct point{ double x; double y;} Point;

typedef struct circle{ Point center; double radius;} Circle;

Example

int is_in_circle(Point *p, Circle *c){ double x_dist, y_dist;

x_dist = p->x - c->center.x; y_dist = p->y - c->center.y;

return (x_dist * x_dist + y_dist * y_dist <= c->radius * c->radius);}

A point is in a circle if its distance from the center is smaller than the radius.

void exit(int status);

Sometimes an error occurs and we want the program to immediately exit (e.g. division by zero, memory allocation failure)

exit terminates the program de-allocates all resources (close open files, free

memory, …) To be used only in extreme situations Remember to #include <stdlib.h>

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