LINKED LIST: Why do I need to "return;" in the if statement...even if i'm passing hea-CodePudding

#include<iostream>
using namespace std;

class node{
    public:
    int data;
    node* addr;.//it's address of the next node
    node(int val){
        data=val;
        addr = NULL;
    }
};

void addVal(node* &head, int val){

    node *n=new node(val);
    if (head==NULL){
        head = n;
        return;///<-WHY THIS?
    }
    node *p=head;
    while(p->addr!=NULL){
        p=p->addr;
    }
    p->addr=n;
    return;
}

void display(node *head){
    node *p=head;
    while(p!=NULL){
        cout<<p<<" "<<p->data<<" "<<p->addr<<endl;
        p=p->addr;
    }
}
int main(){
    node *head=NULL;
    addVal(head,1);
    addVal(head,2);
    addVal(head,3);
    addVal(head,4);
    addVal(head,5);
    display(head);
    return 0;
}

I was getting a blank terminal when I ran this without the return; inside of the if statement. Why is return necessary here? (I passed head's val by reference, so if I change it inside of the function the actual value would be affected as usual. But somehow it gets stuck without the return).

CodePudding user response：

If you do not return this is what happens:

node *p=head;                //   p equals head which equals n
while(p->addr!=NULL){        //   p->addr is NULL, this loop stops immediately
    p=p->addr;
}
p->addr=n;                   //   this assignes n to p's next ptr
return;

The use of many different names for the same thing can be a little confusing. You create a new node and the pointer to it is n. This is the new head. p also points to this new head.

Now addr is the next node of head.

If you do not return after creating the new head and assigning the pointer to head the function will continue to assing the pointer to the same node to head->addr. You'd get a list with a head that has itself as the next node.

You could have observed this by using a debugger.