why is sum a linked list and not a simple integer?

(Toggle Plain Text)

int sum(nd **head){
  nd *pl
  int sum1 = 0;
  p=*head;
  while(p!=NULL)
  {
    sum1  += p->x;
    p=p->next;
  }
    return sum1;
 
}

int sum(nd **head){ nd *pl int sum1 = 0; p=*head; while(p!=NULL) { sum1 += p->x; p=p->next; } return sum1; }

why is sum a linked list and not a simple integer?

(Toggle Plain Text)

int sum(nd **head){
  nd *pl
  int sum1 = 0;
  p=*head;
  while(p!=NULL)
  {
    sum1  += p->x;
    p=p->next;
  }
    return sum1;
 
}

int sum(nd **head){ nd *pl int sum1 = 0; p=*head; while(p!=NULL) { sum1 += p->x; p=p->next; } return sum1; }

actualy i dont know how to di this ...i tired you code it works

how about this one where i want to locate if the linked list is in the lis?

C Syntax (Toggle Plain Text)

 
int locate(nd **head,int num){
  nd *p;
  p=*head;
  while(p!=NULL)
  {
    p=p->next;
    if(num==p->x)
    return 1;
    else
    return 0;
  }
}

int locate(nd **head,int num){ nd *p; p=*head; while(p!=NULL) { p=p->next; if(num==p->x) return 1; else return 0; } }

int locate(nd **head,int num){
nd *p;
p=*head;
while(p!=NULL)
{
p=p->next;
}
if(num==p->x)
return 1;
else
return 0;
}

the code above always return to xero

Look and think about the logic in the code you posted, pay attention to the while loop. All it does is iterate through the linked list without stopping. The if statement needs to be inside the loop -- I did not tell you to move it outside the loop. The else part shoud not be there at all.

C Syntax (Toggle Plain Text)

int locate(nd **head,int num){
  nd *p;
  p=*head;
  while(p!=NULL)
  {
    p=p->next;
    if(num==p->x)
        return 1;
  }
  return 0;
}

int locate(nd **head,int num){ nd *p; p=*head; while(p!=NULL) { p=p->next; if(num==p->x) return 1; } return 0; }