CODEWITHSUNDEEP
c++pointersiteratorlinked-list
SSS
SSS

Reputation: 156

int pointer to the .end() of a list

I am currently working on creating a header file that acts like the standard library. I am in the progress of implemeing .end() for my linked list. In my cpp file I have an integer pointer which is expected to recieve the value of the .end() like

int* itrend;
itrend = ca.end();

ca is just an instance of my object in my header file. In my hpp file, I have a list iterator going to my listofvalues.end(). I would like to return the .end() value like

int* end() 
{
    std::list<value_type>::iterator i = listOfValues.end();
    return (&*i);

}

My "List of values" is just a list holding several ints. eg 

[1,2,3,4,5,6,7,8,9]

In my hpp file, I have a list iterator going to my listofvalues.end(). I would like to return the .end() value (which returns a list iterator); now this posess a problem that it returns the list iterator when I want to have an integer pointer. I have tried to deference it and take the address by using &* and this will give an error saying "List Iterator Not Dereferencable". I know it's because I'm trying to get the element after the last value, but that is exactly what I need to return, since I am trying to re-implement the linked list .end() method.

Does anyone know how I can have an integer pointer point to the List.end()?

Any assistance is appreciated.

Upvotes: 2

Views: 2023

Answers (4)

marcinj
marcinj

Reputation: 49976

how do you use your own impementaton of end() anyway? why not returning null when iterator whould return end() ?

int* end() 
{
    return NULL;
}

Upvotes: 2

111111
111111

Reputation: 16148

The idea of representing the end of the a Linked list with a pointer is somewhat flawed, because you can't progress a pointer with ++ in a linked list, it doesn't iterate through anything.

If you you want to get the back of a list implement l.back() and return a reference to the last value.

Upvotes: 0

hmjd
hmjd

Reputation: 121961

To implement begin() and end() in a style similar to STL return iterators, not pointers:

template <typename value_type>
class my_list
{
public:
    typedef std::list<value_type>::iterator iterator;
    typedef std::list<value_type>::const_iterator const_iterator;

    iterator begin() { return listofValues.begin(); }
    const_iterator begin() const { return listofValues.begin(); }


    iterator end() { return listofValues.end(); }
    const_iterator end() const { return listofValues.end(); }
};

Upvotes: 4

Kerrek SB
Kerrek SB

Reputation: 476980

You cannot. That's why the standard library containers are more general than a flat array. In a C-style array, you can form the pointer one-past-the end (although you are not allowed to dereference it) and use it as a sentinel. In a more abstract container, there is no such "natural" mechanic, which is why iterators were invented in the first place: They are a generalization of pointers that works in more general contexts. Each container's iterator type must solve the problem of providing a suitable "end" sentinel value in its own way.

Upvotes: 2

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