C++ list functions not worling

user_18881624 0
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I'm tryin to create a program that let me add elements in tail of a list and then prints them. It doesn't give me an error but he doesn't do anything. What am I doing wrong ?

#include<iostream>
using namespace std;
 
struct lista{
    int val;
    lista *next;
};

typedef lista* ptr_lista;

void tail_add(ptr_lista head, int valore){
    if(head=NULL){
         head=new lista;
         head->val=valore;
         head->next=NULL;
    } else {
        ptr_lista p=head;
        while(p->next!=NULL){
                p=p->next;
        }
        p->next=new lista;
        p->next->val=valore;
        p->next->next=NULL;
    }
}

void print(ptr_lista p){
     while(p!=NULL){
            cout<<p->val<< " ";
            p=p->next;
      }
}

int main(){
    ptr_lista m;
    tail_add(m,5);
    tail_add(m,6);
    print(m);
}
singly-linked-list pass-by-value pass-by-reference struct c++
1 Answers

For starters the pointer m is not initialized and has an indeterminate value

ptr_lista m;

You need to initialize it

ptr_lista m = nullptr;

The function accepts the pointer by value

void tail_add(ptr_lista head, int valore){

So changing the parameter head within the function like

 head=new lista;

has no effect on the original pointer m declared in main.

You need to declare the parameter as a reference to the pointer

void tail_add(ptr_lista &head, int valore){

The function can be defined the following way

void tail_add( ptr_lista &head, int valore )
{
    ptr_lista new_lista_ptr = new lista { valore, nullptr };

    if ( head == nullptr )
    {
         head = new_lista_ptr;
    } 
    else 
    {
        ptr_lista p = head;

        while ( p->next ) p = p->next;

        p->next = new_lista_ptr;
    }
}

Pay attention to that it is not a good idea to introduce an alias for a pointer like

typedef lista* ptr_lista;

For example if you will write

const ptr_lista

then it means

lista * const

not

const lista *

that is required for the declaration of the parameter of the function print because it does not change nodes of the list.

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