Reiterating a linked list (Bankers Algorithm)

Question

I am making a simple bankers algorithm simulator. When I compare the need with the available resources it works fine for 1 loop. However I cannot get it to reiterate the linked list again. (In bankers algorithm you may only be able to run the last in the linked list. In that case you would have to go through the linked list again to see if anymore can run[this is the part that is not working])I think it has something to do with the pointers, but I'm not sure what.

struct LL //linked list structure(pcb)
{
    LL_pid pid;
    int alloc[15];
    int max[15];
    int need[15];
    int finish;//flag to show if finished
    PCB *next; //points to next pcb in linked list
};

Heres where I am stumped. I added some test printfs and realized that it does not re iterate the loop(probably since the pointer of pcb_head is now null)?

void makeBANK(PCB *pcb_head, int avail[15]){
    PCB *temp=pcb_head;
    int availnew[15];
    int x=0;
    for(x=0;x<processCount;x++){//get all the available resources
        availnew[x]=avail[x];
    }
    int alldone=0;//check if all processes could run
    int possible=0;//check if its possible if a process can run with current available resources
    int y=0;
    int i=0;
    int z=0;
    for(y=0;y<processCount;y++){//trying to iterate the linked list at least the amount of processes there are(worst case)
        temp=pcb_head;
        while((temp!=NULL)&&(temp->finish!=1)){//search all nodes
            for(i=0;i<resourceCount;i++){//compare avail and need
                if(availnew[i]>=temp->need[i]){//if possible keep 1 ,loop all
                    possible=1;
                }
                else{
                    possible=0;
                    printf("oops");
                    break;
                }//if not possible break
            }
            if(possible==1){//if the possible still 1 then print
                printf("%d running",temp->pid);
                temp->finish=1;
                alldone++;
                for(z=0;z<resourceCount;z++){//add the allocated to the available
                    availnew[z]=availnew[z]+temp->alloc[z];
                    printf("avil: %d",availnew[z]);
                }
            }
            temp=temp->next;//and go to next node(also needed for else)
        }
    }
    if(alldone!=processCount)
        printf("not safe");
    else
        printf("safe");
}

I am open to all tips(organization..etc) even if I get downvoted for a possibly simple solution I could not find on google.

Answer 1

All right, the while loop needs to terminate when either it has reached end of linked list (temp != NULL) or when the current process is finished (temp -> finish = 1).

Problem is what if there is a process in linked list which has already finished?

I can suggest two solutions:

1. remove the task from current linked list: if you are not going to need the task it could be deleted from linked list. But this may become quite complex.
2. use a different flag variable: a flag variable that tells whether in current iteration of while loop a particular process(any process) gets finished or not.

this is demo code for the situation we have :

current condition:

for(i = 0; i < numberOfNodes; i++) {
    temp = listHead; 
    while(temp != NULL && temp -> finish != 1) {
        //check resource availability here 
        if(executionPossible = 1) { // Process can be executed
            //print process id here

            temp -> finish = 1;

            //update number of available resources
        }
        temp = temp -> next; 
    }
}

My Solution(second one) :

for(i = 0; i < numberOfNodes; i++) {
    int currentFinished = 0;
    temp = listHead; 
    while(temp != NULL && currentFinished != 1) {
        //check resource availability here 
        if(executionPossible = 1) { // Process can be executed
            //print process id here and mark/set process as finished
            temp -> finish = 1;
            currentFinished = 1;

            //update number of available resources
        }
        temp = temp -> next; 
    }
}

PS : This is my first answer on stack overflow. Feedback are welcome.