Thread: Custom Template

The Linked List should know nothing about object's datamembers. For the display() method to work, the object must overload ostream operator <<, otherwise the template won't compile.

In other words, implementing this operator is one of the requirements for a type that can be stored in this list.

Actually, the way templates work, it's only a requirement if you actually call display(). You can store objects in the list that do not overload << as long as you don't call display() for that particular list.

Originally Posted by CornedBee

Actually, the way templates work, it's only a requirement if you actually call display(). You can store objects in the list that do not overload << as long as you don't call display() for that particular list.

Ah, that I didn't know, thinking that all methods need to be compilable if a templated class is instantiated.

If possible, what do you recommend me to change? The private variable of *first to public? I understand that would ruin the concept of data hiding because I'm quite raw in the operator overloading.

Normally, displaying the contents is not the kind of task you'd expect from a List. A list's main task is to hold data.

The list should provide some means to iterate over its contents, so the user can access the contents of each node and deside what to do with it.

Supposing the following class:

Code:

class MyNumber
{
    public:
        MyNumber(int n): value(n) {}
        void print() const {std::cout << value << '\n';}
    private:
        int value;
};

If you stored those in a std::list, that would give you iterators that you can use to access each MyNumber in the list and call the print method.

Code:

//create a std::list with some data
std::list<MyNumber> myList;
for (int i = 0; i < 10; ++i) {
    myList.push_back(i);
}

//Now iterate over the list and print each number
for (std::list<MyNumber>::iterator it = myList.begin(); it != myList.end(); ++it) {
    it->print();
}

Note, that the iterator does not expose the list itself (for example a head pointer). It only lets you move through the list (from begin() to end()) and access each MyNumber that you have stored in it.

Writing your own iterators is a bit of a black magic (there's a recent thread discussing that), so you might stay with your display() method and overload the stream << operator for MyNumber. This is done like that:

Code:

class MyNumber
{
    public:
        MyNumber(int n): value(n) {}
        friend std::ostream& operator<< (std::ostream& os, const MyNumber& num);
    private:
        int value;
};

std::ostream& operator<< (std::ostream& os, const MyNumber& num)
{
    return os << num.value;
}

The overload is a free function which is declared to be a friend of MyNumber, so it can access the otherwise private value member. Now MyNumber is "coutable":

Code:

MyNumber a(12), b(32);
std::cout << "First: " << a << ", second: " << b << std::endl;

The overload need not (and thus should not) be a friend if all data it wants to display can be accessed through the public interface of the class.

I.e. if MyNumber has a method

Code:

public: int get_value() const { return value; }

then there's no need for operator << to be a friend. It can be implemented thusly:

Code:

std::ostream & operator <<(std::ostream &os, const MyNumber &num)
{
  return os << num.get_value();
}

Thanks for the advice but I'm sorry, I find it this quite hard to swallow. My current work now is to analyze various data storages and std::list is one of them.

My object classes are derived from an abstract base class.

My current task was to create a custom template list along with other data structures. I do have accessors in my classes.

Correct me if I'm wrong, so if I put the operator overloading in my seperate classes, wouldn't that overwrite other random std::cout << statements I placed for testing purposes?

*EDIT
or if I specify std::cout << "First: " << a << "\n"; and "a" is an object from MyNumber it will use the overload operator.

The operator will not be overloaded if I just put std::cout << "Test\n";

The overload only tells cout how to print your particular type and won't change the way other types are cout'ed.

This bitshift operator is overloaded for ostream in the first place, and there already are multiple overloads.

Operator overloading is basically the same thing as function overloading :

Code:

#include <iostream>

void foo(double)
{
    std::cout << "Double\n";
}

void foo(int)
{
    std::cout << "Int\n";
}

int main()
{
    foo(23); //calls foo(int)
    foo(2.3); //calls foo(double)
}

So do I put the operator overloading in the base class or on the derived classes?

*EDIT
Apparently, it clashed with my other display functions from the same objects, or so I thought.

Code:

AsteroidList.o:AsteroidList.cpp:(.text+0x340): first defined here
gameboard.o: In function `operator<<(std::basic_ostream<char, std::char_traits<char> >&, AsteroidType1)':

Here's a piece of it but there more than just that.

Sounds like you implemented a non-inline non-template function in a header file. You mustn't do that. Put all function bodies that are neither inline nor templates in .cpp files.

Declare it in the header, implement it in a source file.

Well, actually I'd make the << operator inline. But laserlight's method is the best one for larger functions. Nothing stops you from adding another source file, even if the class it is named after is abstract.