我应该如何用所有不同的链表设置格式的命名空间？

Question

I want make a very simple namespace 'list' kind of like 'std' with libraries like 'singly' and 'doubly'. How can I modular the code so that I can only use variables I need to (circular vs non-circular)?

I imagine that I can just add a 'struct node* prev' to the struct that the list is using, but if the user decides to not do circular, it would always be 'nullptr' causing in an unnecessary variable. The second way I imagine is with having 2 different classes...

namespace list{
    struct snode{
        int data;
        struct snode* next;
    };

    class singly{
        public:
            singly();
            singly(unsigned long long amount=0, bool circular=0, bool userCreated=0, bool empty=0);
            ~singly();

            void create(unsigned long long amount=0, bool circular=0, bool userCreated=0, bool empty=0);
            void display();
            void destroy();
        private:
            snode* HEAD;
            unsigned long long amount;

            bool empty;
            bool circular;
            snode* TAIL;
    };
}

I expect with this namespace to do the following..

include "singly.h"

using namespace list;

singly list;

list.create(5);

list.display();

list.destroy();

or by not using the namespace something else...

list::singly list(7, 1, 0, 1);

list.display();

list.~list();

While this isn't an assignment or anything, I want this to be my go-to when I need to create a linked list ever in my life. Eventually, I want to create a namespace 'tree' and have further expansion with that. I almost want this to be as easy to use as the 'string' abstract data type.

Answer 1

short answer

It's impossible

long answer

I lied(ish). Create a base strict containing data, then create snode that inherits the data, implementing the pointer of base class. When going through the list check whether it fails.

struct A {
    int data;
    // make it polymorphic for the conversion
    virtual void function() {return;}
};

struct B : A {
    struct A* next;
};

int main()
{
    struct A end;
    struct B beginning;
    beginning.next = &end;
    struct B* buffer = &beginning;
    while(true) {
        buffer = dynamic_cast<struct B*>(buffer->next);
        if (buffer == nullptr) break;
    }
}

why it's pretty irrelevant

You only save at most (0 bytes) of memory per list. You save it using struct A, but struct A effectively has a function ptr/address aswell, which takes up space. This, however may be fixed by the compiler. And it takes longer to write worse code and the overall efficiency will be worse.