Uploaded Dining Philosopher problem solution

C/Dining_Philosopher.c

@@ -0,0 +1,91 @@
+#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;
+}