linked list pass by address error

[corbett]

#include <iostream>
using namespace std;

struct node
{
	int data;
	node*next;
};

void add(node**List, int number);

void display(node*List);

int main()
{
	cout<<"\n1. Add\n2.Search\n3.Delete\n4.Display\nWhat do you want to do? \n";
	int choice;
	cin>>choice;

	node*List = NULL;

	switch (choice)
	{
		case 1:
		cout<<"Enter number to add! :";
		int number;
		cin>>number;
		add(&List,number);
		
		main();
		break;

		case 2:
		cout<<"Enter number to search! :";
		cin>>number;
		//search(List, number);
		main();
		break;

		case 3:
		cout<<"Enter number to delete! :";
		cin>>number;
		//delete(&List, number);
		main();
		break;

		case 4:
		display(List);
		main();
		break;
	}
}

void add(node**List, int number)
{
	if((*List) == NULL)
	{
		
		(*List) -> data = number;
		(*List) -> next = NULL;
		cout<<(*List)->data;

	}
	else
	{
		node*List2 = *List;
		while(List2->next != NULL)
		{
			List2 = List2->next;
		}
		List2->next = new node;
		List2->next->data = number;
		List2->next->next = NULL;
	}
}

void display(node*List)
{
	cout<<"The contents of the linked list are: \n";
	while(List != NULL)
	{
		cout<<List->data<<endl;
		List=List->next;
	}
}

How is this not working? just the add & display function.

[abhishekm71]

Is there any specific reason why you are complicating your code by using pointer to pointer instead of a simple pointer to node?

Also, one thing that I note is that you are not allocating memory for the first node.

if((*List) == NULL)
	{
	        // *List = new node;
		(*List) -> data = number;
		(*List) -> next = NULL;
		cout<<(*List)->data;

	}

Don't forget to delete your 'newed' nodes at the end of your program.

[corbett]

if((*List) == NULL)
	{
	        // *List = new node;
		(*List) -> data = number;
		(*List) -> next = NULL;
		cout<<(*List)->data;

	}

Thanks for this.

But still, the values added to the linked list is not displayed with the display function (or even in the main function if tried).

[corbett]

Don't forget to delete your 'newed' nodes at the end of your program.

How do i do this?

[lsk]

You shouldn't call many main function, it's no point.
It'll be like you're starting new program with old GLOBAL VARIABLES but new ALL MAIN'S LOCAL VARIABLES.(And it make your List = NULL every times)
If you want to run as loop, use loop.

#include <iostream>
using namespace std;

struct node
{
	int data;
	node*next;
};

void add(node**List, int number);

void display(node*List);

int main()
{

	node*List = NULL;

	while(true){
		
		cout<<"\n1. Add\n2.Search\n3.Delete\n4.Display\nWhat do you want to do? \n";
		int choice;
		cin>>choice;
		switch (choice)
		{
			case 1:
			cout<<"Enter number to add! :";
			int number;
			cin>>number;
			add(&List,number);
			
			//main();
			break;
	
			case 2:
			cout<<"Enter number to search! :";
			cin>>number;
			//search(List, number);
			//main();
			break;
	
			case 3:
			cout<<"Enter number to delete! :";
			cin>>number;
			//delete(&List, number);
			//main();
			break;
	
			case 4:
			display(List);
			//main();
			break;
			default:
			break;
		}
	}
}

void add(node**List, int number)
{
	if((*List) == NULL)
	{
		(*List) = new node;
		(*List) -> data = number;
		(*List) -> next = NULL;
		cout<<(*List)->data;

	}
	else
	{
		node*List2 = *List;
		while(List2->next != NULL)
		{
			List2 = List2->next;
		}
		List2->next = new node;
		List2->next->data = number;
		List2->next->next = NULL;
	}
}

void display(node*List)
{
	cout<<"The contents of the linked list are: \n";
	while(List != NULL)
	{
		cout<<List->data<<endl;
		List=List->next;
	}
}

[lsk]

To delete all nodes you created.
Loop by your List and use delete from tail.

node *nn;
nn = new node;
delete nn;

Do it for preventing memory leak.