Can't figure out the pointer logic in this example

Question

I can't figure this out for the life of me. This is a linked list example.

I'm a beginner. I've been looking at this for about an hour and I'm ready to fold and look for assistance. Heh.

So basically the issue is that there is a pointer on one of the functions, and pointers inside the functions, and pointers bloody everywhere and I can't seem to figure out the series of logical steps that are being taken here.

What I did finally figure out is that Entry *newOne is defining a "struct Entry" pointer. I don't really get what is happening in the full statement, or how the different parts are calling values. At all.

More specifics below.

#include <iostream>
#include <vector>
#include <fstream>
#include <string>
#include <sstream>
using namespace std;

struct Entry {
    string name, phone;
    Entry *next;
};

void PrintEntry(Entry *e)
{
    cout << e->name << " " << e->phone << endl;
}

Entry *GetNewEntry()
    {       
        Entry *newOne = new Entry;
        cout << "Enter name (ENTER to quit):";
        string name;
        getline (cin,name);
        if (name == "") return NULL;
        newOne->name = name;
        cout << "Enter phone: ";
        string phone;
        getline(cin, phone);
        newOne->phone = phone;
        newOne->next = NULL; // no one follows
        return newOne;
    }
int main () {
    Entry *n = GetNewEntry();
    PrintEntry(n);
    return 0;
}

1. Entry *newOne = new Entry (I don't understand this - isn't new Entry just an address for a struct Entry? and isn't Entry *newOne a pointer? Then isn't this just assigning the value of the pointer to an address... quite lost.

And if newOne is simply an address (which it is after verifying) then why does saying newOne->phone=phone do anything? That doesn't make sense!

2. Entry *GetNewEntry() (I don't understand this - at the end of the function an address to the newOne entry is returned - does the * add to this "return" value perhaps)

3. Entry *n = GetNewEntry() (With relation to the function having the pointer symbol on it - GetNewEntry either returns the newOne memory address, or the newOne pointer - and Entry *n being a struct Entry pointer would then be set either to that memory address (much like Entry *newOne = new Entry) or it would be set to the pointer to that address... ugh)

4. PrintEntry(n) refers back to PrintEntry(Entry *e)

As you can see I'm confused.

Answer 1

The first thing in this code:

struct Entry {
string name, phone;
Entry *next;};

is a linked list of possible phone book entries. I hope you are familiar with the linked list otherwise check here http://www.cprogramming.com/tutorial/lesson15.html

If you trace the code in main it is very simple. First of all it creates an Entry by calling GetNewEntry function. This function prompts the user for data like name and phone number and create the Entry object.

It sets the "next" variable in the structure as null because probably this is the only element it is going to create and returns the created object back to the main function.

Now the PrintEntry function will print the data inside the Entry that just has been created. From my point of view the "next" variable in the Entry struct is useless here but maybe in your lesson there is going to be a LinkedList which make use of this variable.

Answer 2

1. Entry *newOne = new Entry; (I don't understand this - isn't new Entry just an address for a struct Entry? and isn't Entry *newOne a pointer? Then isn't this just assigning the value of the pointer to an address... quite lost.

A pointer is a variable that holds an address. new Entry creates a new Entry on the heap, and then returns its address. Then this address is stored in newOne. This line also declares newOne, but it could have been written separately as Entry *newOne; newOne = new Entry;.

And if newOne is simply an address (which it is after verifying) then why does saying newOne->phone=phone do anything? That doesn't make sense!

Why wouldn't it? newOne->phone is short for (*newOne).phone. *newOne means "the thing whose address is newOne", so it's the Entry that we just created a moment ago (with new Entry). So (*newOne).phone is that Entry's member called phone.

2. Entry *GetNewEntry() (I don't understand this - at the end of the function an address to the newOne entry is returned - does the * add to this "return" value perhaps)

Entry * is the return type. It could alternatively be written as Entry* GetNewEntry(). This function returns a pointer to an Entry (or the address of an Entry; call it whichever you want). * in a type means "pointer to" (or "address of") the thing before it.

3. Entry *n = GetNewEntry(); (With relation to the function having the pointer symbol on it - GetNewEntry either returns the newOne memory address, or the newOne pointer - and Entry *n being a struct Entry pointer would then be set either to that memory address (much like Entry *newOne = new Entry) or it would be set to the pointer to that address... ugh)

This means the same as Entry *n; n = GetNewEntry();. It declares the variable n (with type Entry*). Then it calls GetNewEntry(). Then it sets n to the return value from GetNewEntry()

4. PrintEntry(n) refers back to PrintEntry(Entry *e)

This is a function call. I'm not sure what you're asking here.