DinningPhilosopher.c

// Dinning Philosopher problem

#include<stdio.h>
 
#define n 4
 
int compltedPhilo = 0, i;
 
struct fork{
    int taken;
}ForkAvil[n];
 
struct philosp{
    int left;
    int right;
}Philostatus[n];
 
void goForDinner(int philID){ //same like threads concept here cases implemented
    if(Philostatus[philID].left==10 && Philostatus[philID].right==10) {
        printf("Philosopher %d completed his dinner\n",philID+1);
    //if already completed dinner
    } else if(Philostatus[philID].left==1 && Philostatus[philID].right==1){
        //if just taken two forks
        printf("Philosopher %d completed his dinner\n",philID+1);

        Philostatus[philID].left = Philostatus[philID].right = 10; //remembering that he completed dinner by assigning value 10
        int otherFork = philID-1;

        if(otherFork== -1)
            otherFork=(n-1);

        ForkAvil[philID].taken = ForkAvil[otherFork].taken = 0; //releasing forks
        printf("Philosopher %d released fork %d and fork %d\n",philID+1,philID+1,otherFork+1);
        compltedPhilo++;
    } else if(Philostatus[philID].left==1 && Philostatus[philID].right==0){ //left already taken, trying for right fork
        if(philID==(n-1)){
            if(ForkAvil[philID].taken==0){ //KEY POINT OF THIS PROBLEM, THAT LAST PHILOSOPHER TRYING IN reverse DIRECTION
                ForkAvil[philID].taken = Philostatus[philID].right = 1;
                printf("Fork %d taken by philosopher %d\n",philID+1,philID+1);
            }else{
                printf("Philosopher %d is waiting for fork %d\n",philID+1,philID+1);
            }
        }else{ //except last philosopher case
            int dupphilID = philID;
            philID-=1;

            if(philID== -1)
                philID=(n-1);

            if(ForkAvil[philID].taken == 0){
                ForkAvil[philID].taken = Philostatus[dupphilID].right = 1;
                printf("Fork %d taken by Philosopher %d\n",philID+1,dupphilID+1);
            }else{
                printf("Philosopher %d is waiting for Fork %d\n",dupphilID+1,philID+1);
            }
        }
    } else if(Philostatus[philID].left==0){ //nothing taken yet
            if(philID==(n-1)){
                if(ForkAvil[philID-1].taken==0){ //KEY POINT OF THIS PROBLEM, THAT LAST PHILOSOPHER TRYING IN reverse DIRECTION
                    ForkAvil[philID-1].taken = Philostatus[philID].left = 1;
                    printf("Fork %d taken by philosopher %d\n",philID,philID+1);
                }else{
                    printf("Philosopher %d is waiting for fork %d\n",philID+1,philID);
                }
            }else{ //except last philosopher case
                if(ForkAvil[philID].taken == 0){
                    ForkAvil[philID].taken = Philostatus[philID].left = 1;
                    printf("Fork %d taken by Philosopher %d\n",philID+1,philID+1);
                }else{
                    printf("Philosopher %d is waiting for Fork %d\n",philID+1,philID+1);
                }
            }
    }else{}
}
 
int main(){
    for(i=0;i<n;i++) {
        ForkAvil[i].taken=Philostatus[i].left=Philostatus[i].right=0;
    }
    while(compltedPhilo<n){
        /* Observe here carefully, while loop will run until all philosophers complete dinner
        Actually problem of deadlock occur only thy try to take at same time
        This for loop will say that they are trying at same time. And remaining status will print by go for dinner function
        */
        for(i=0;i<n;i++)
            goForDinner(i);
        printf("\nTill now num of philosophers completed dinner are %d\n\n",compltedPhilo);
    }

    return 0;
}