C Program to solve Dining philosophers problem.

Write a C Program to solve Dining philosophers problem.
Dining philosophers problem is a classic synchronization problem.A problem introduced by Dijkstra concerning resource allocation between processes. Five silent philosophers sit  around table with a bowl of spaghetti. A fork is placed between each pair of adjacent philosophers.

Each philosopher must alternately think and eat.
Eating is not limited by the amount of spaghetti left: assume an infinite supply.
However, a philosopher can only eat while holding both the fork to the left and the fork to the right
(an alternative problem formulation uses rice and chopsticks instead of spaghetti and forks).

Each philosopher can pick up an adjacent fork, when available, and put it down, when holding it.
These are separate actions: forks must be picked up and put down one by one.
The problem is how to design a discipline of behavior (a concurrent algorithm) such that each philosopher won’t starve, i.e. can forever continue to alternate between eating and thinking.
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#include<stdio.h>
#include<semaphore.h>
#include<pthread.h>

#define N 5
#define THINKING 0
#define HUNGRY 1
#define EATING 2
#define LEFT (ph_num+4)%N
#define RIGHT (ph_num+1)%N

sem_t mutex;
sem_t S[N];

void * philospher(void *num);
void take_fork(int);
void put_fork(int);
void test(int);

int state[N];
int phil_num[N]={0,1,2,3,4};

int main()
{
int i;
pthread_t thread_id[N];
sem_init(&mutex,0,1);
for(i=0;i<N;i++)
sem_init(&S[i],0,0);
for(i=0;i<N;i++)
{
pthread_create(&thread_id[i],NULL,philospher,&phil_num[i]);
printf("Philosopher %d is thinkingn",i+1);
}
for(i=0;i<N;i++)
pthread_join(thread_id[i],NULL);
}

void *philospher(void *num)
{
while(1)
{
int *i = num;
sleep(1);
take_fork(*i);
sleep(0);
put_fork(*i);
}
}

void take_fork(int ph_num)
{
sem_wait(&mutex);
state[ph_num] = HUNGRY;
printf("Philosopher %d is Hungryn",ph_num+1);
test(ph_num);
sem_post(&mutex);
sem_wait(&S[ph_num]);
sleep(1);
}

void test(int ph_num)
{
if (state[ph_num] == HUNGRY && state[LEFT] != EATING && state[RIGHT] != EATING)
{
state[ph_num] = EATING;
sleep(2);
printf("Philosopher %d takes fork %d and %dn",ph_num+1,LEFT+1,ph_num+1);
printf("Philosopher %d is Eatingn",ph_num+1);
sem_post(&S[ph_num]);
}
}

void put_fork(int ph_num)
{
sem_wait(&mutex);
state[ph_num] = THINKING;
printf("Philosopher %d putting fork %d and %d downn",ph_num+1,LEFT+1,ph_num+1);
printf("Philosopher %d is thinkingn",ph_num+1);
test(LEFT);
test(RIGHT);
sem_post(&mutex);
}
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C Program to demonstrate dynamic memory allocation example.

Write a C Program to demonstrate dynamic memory allocation example.
Dynamic memory allocation means you can allocate or relocate (manipulate) the memory at the run time, using malloc, calloc, and realloc functions.
Using malloc, We can allocate block of memory for a variable
Using calloc function, We can allocate multiple blocks of memory for a variable.
We can alter, reassign the allocated memory using the realloc function.
We can releasing the allocated memory using the free function. Read more about C Programming Language .
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* programs for commercial purposes,
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*
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***********************************************************/



#include<stdio.h>
#include<stdlib.h>

int main() {
int* grades;
int sum = 0, i, numberOfStudents;
float average;


printf("Enter the number of students: ");
scanf("%d", &numberOfStudents);
getchar();

if((grades = (int*) malloc(numberOfStudents * sizeof(int))) == NULL) {
printf("nError: Not enough memory to allocate grades arrayn");
exit(1);
}

printf("nEnter the grades of %d students (in separate lines):n", numberOfStudents);

for(i = 0; i < numberOfStudents; i++) {
scanf("%d", &grades[i]);
getchar();
}

/* calculate sum */
for(i = 0; i < numberOfStudents; i++)
sum += grades[i];

/* calculate the average */
average = (float) sum / numberOfStudents;

printf("nThe average of the grades of all students is %.2f",
average);

getchar();
return(0);
}

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C Program to demonstrate isalpha, isdigit, is space.

C Program to demonstrate the following functions:isalpha, isdigit, isspace.The same principles apply to isalnum, iscntrl, isgraph,islower, isprint, ispunct, isupper, isxdigit.
In the standard library ctype.h all the above functions are declared.
All the subroutines mentioned here, returns non zero(true), If the checked argument is true for the respective subroutines. Read more about C Programming Language .
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* You can use all the programs on www.c-program-example.com
* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact info@c-program-example.com
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

#include <stdio.h> /* printf */
#include <ctype.h> /* isalpha isdigit isspace etc */

#define FALSE 0
#define TRUE 1

/* function declarations */
int char_type(char);

main()
{
char ch;
/* get a character from the keyboard */
printf(" Please enter a charcater => ");
ch = getc(stdin);

char_type(ch); /* Figure out the character type */


}

//char_type:decides the character type
int char_type(char ch)
{
/* returns non zero if A-Z or a-z */
if ( isalpha(ch) != FALSE)
printf("%c is an Alpha character.n",ch);

/* returns non zero if 0-9 */
if ( isdigit(ch) != FALSE)
printf("%c is a numeric character.n",ch);

/* returns non zero if a space, CR, Tab, NL FF */
if ( isspace(ch) != FALSE)
printf("%c is white spacen", ch);

}

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K & R C Programs Exercise 7-2.

K and R C, Solution to Exercise 7-2:
K and R C Programs Exercises provides the solution to all the exercises in the C Programming Language (2nd Edition). You can learn and solve K&R C Programs Exercise.
Write a program that will print arbitrary input in a sensible way. As a minimum, it should print non-graphic characters in octal or hexadecimal according to local custom, and break long text lines. Read more about C Programming Language .
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***********************************************************/

#include <stdio.h>
#define OCTAL 8
#define HEXADECIMAL 16


void ProcessArgs(int argc, char *argv[], int *output)
{
int i = 0;
while(argc > 1)
{
--argc;
if(argv[argc][0] == '-')
{
i = 1;
while(argv[argc][i] != '