Question

I’m trying to swap a pointer but all I get is access errors, are there any way to do this?

void Swap(someObject *first, someObject *next)
{
     delete first;
     first = next;
     // I'm guessing this delete first pointer as well ?
     delete next;
     next = new someObject();
};

Clarification on the method:
It should just swap first with second and create a new instance of “someObject” on second, it shouldn’t swap like = FIRST = SECOND, SECOND = FIRST. Just first becomes second, and second becomes a new object or a new pointer to an object.

Second = Next.

SOLUTION

void Swap(someObject*& first, someObject*& next)
{
    std::swap(first, next);
    next = new someObject();
};

Answer 1

Try this swap idiom instead. Yours modifies the parameters (and deletes both objects) but does not actually swap the pointer values at the call site.

template <typename T>
void Swap(T & a, T & b)
{
    T x = a;

    a = b;
    b = x;
}

However, you could just as easily use std::swap for this; it will effectively do the same thing this template function does.