Computers and programming 1

The 11th lecture

Jiří Šebesta

TOPIC – programming with dynamic variables

1. Coupling with dynamic variables2. Dynamic database - example

Coupling with dynamic variables (1/13)

• Approach using array of pointers

Allocated static array of pointer to dynamic variable (structure) – max. number of structures is fixed in advance

Coupling with dynamic variables (2/13)

• Linear list approach

The first pointer to the first structure is only allocated in memory. Each structure has pointer to the following structure as item – max. number of structures is restricted only by memory capacity.

Coupling with dynamic variables (3/13)

• Linear list approach - modified

Two static pointers to the first and last structure (for simple implementation).

Coupling with dynamic variables (4/13)

typedef struct comp // competitor record{

char name[10];int jump;struct comp *next;

}t_comp; t_comp *first; // ptr to the first comp. - globalt_comp *last; // ptr to the last comp. - globalint cnt = 0; // the number of recorded comp. - global

• Linear list – insertion of unknown number of competitors (broad jumpers) in to the list including results (length of jump)• Structure and global variables

comp is structure namespace

t_comp is type namespace

Coupling with dynamic variables (5/13)

void add(char *cname, int cjump) // add comp. record{

t_comp *ccomp; // ptr to a compccomp = (t_comp*)malloc(sizeof(t_comp));strcpy(ccomp->name, cname); // record fillingccomp->jump = cjump; cnt++; // the number of records plus oneif(cnt == 1){ first = ccomp; // add the first record

last = ccomp;}else{ last->next = ccomp; // add next record

last = ccomp;}

}

• Function for adding of a competitor and his length of jump to the linear list

Coupling with dynamic variables (6/13)

void show(void){

t_comp *scomp;int acnt=cnt;

scomp = first;do{ printf("%s: %d cm\n", scomp->name, scomp->jump); scomp=scomp->next;}while (--acnt > 0);

}

• List of competitors printing including length of jump according to insertion order (from the first to the last)

Coupling with dynamic variables (7/13)

int main(void){

char cmd, aname[10];int ajump;

printf("\nA: Add, S: Show, Q: Quit\n");scanf("%c", &cmd); fflush(stdin);while(!(cmd == 'Q' || cmd == 'q')){

if(cmd=='A' || cmd=='a'){

printf("\nName: ");scanf("%s", aname); fflush(stdin);printf("\nJump [cm]: ");scanf("%d", &ajump); fflush(stdin);add(aname, ajump);

}

• main() function for application with commands

Coupling with dynamic variables (8/13)

if(cmd=='S' || cmd=='s')show();

printf("\nA: Add, S: Show, Q: Quit");scanf("%c", &cmd); fflush(stdin);

}return 0;

}

Example: Ex71.c

Coupling with dynamic variables (9/13)

• Modification of the previous example by algorithm for searching of the best competitor

• Replenishment of the function add()

void add(char *cname, int cjump) // add comp. record{

// the same as in the example 71if(cnt == 1){ first = ccomp; // add the first record

last = ccomp;}else{ last->next = ccomp; // add next record

last = ccomp;}last->next = NULL; // last points to null address

}

Coupling with dynamic variables (10/13)

void results(void){ t_comp *scomp, *gold; int mjump = first->jump; gold = first; scomp = first->next; do {

if(scomp->jump > mjump) { gold = scomp; mjump = gold->jump; }

scomp = scomp->next; } while(scomp != NULL); printf("\nWin %s (%d cm)\n", gold->name, gold->jump);}

Example: Ex72.c

Coupling with dynamic variables (11/13)

• Stack approach

One static pointer to the top of stack (heap) – system LIFO (last in – first out).

Coupling with dynamic variables (12/13)

• Tree approach

Coupling with dynamic variables (13/13)

• Complicated coupling of structures – family tree

strcpy(me->p_sists[0]-> Name, ”Jana”);

me->p_sists[0]-> p_mother = me-> p_mother;

//uncles:me->p_mother-> p_brths[x]me->p_father-> p_brths[x]

Dynamic database - example (1/5)

• Submission:• Build-up a console application, which dynamically generate a database of competitors (broad jumpers). Each structure consists of competitor’s name, competitor’s country, length of the jump, and pointer to the other structure of competitor. Create a function for appending of the competitor including length of jump and function for printing of the competitors including their results. • Modify the function for appending of competitor: a new competitor have to be add to the linear list according to his result (length of the jump) – list is always sorted regarding the length of the jump.

Dynamic database - example (2/5)

• Linear sorted list

Dynamic database - example (3/5)

typedef struct t_comp // competitor record{

char name[10];char country[10];int jump;t_comp *next;

};

t_comp *first; // ptr to the first comp. - globalt_comp *last; // ptr to the last comp. - globalint count = 0; // the number of recorded comp. - global

• Linear list – insertion of unknown number of competitors (broad jumpers) to the list including result (length of the jump)

• Structure and global variables

Dynamic database - example (4/5)

void add(char *cname, char *ccountry, int cjump) {

t_comp *ccomp, *prevcomp, *nextcomp; ...

• Modified function for appending of the competitor and his length of the jump to the linear list with automatic sorting regarding the result

• Programming in lecture

Dynamic database - example (5/5)

Example: Ex73.c

Complete application will be issued in web pages in December, 5 as Ex73sol.c

Records releasing from memory

TOPIC OF THE NEXT LECTURE

1. Advanced algorithms in C

THANK YOU FOR YOUR ATTENTION