Question

I know that you can take a pointer such as:

someFunction(const char* &txt)
{
    txt++; //Increment the pointer
    txt--; //Decrement the pointer
}

how do I get back to the beginning of the pointer? (without counting my number of increments and decrementing that amount) I was thinking about something like:

txt = &(*txt[0]);

But that isn’t seeming to work (assigning the pointer to its beginning location).

Any help would be greatly appreciated

Brett

Answer 1

You can’t. You have to either save it’s original position:

const char *o = txt;
// do work
txt = o;

Or use an index:

size_t i = 0;
// instead of txt++ it's i++
// and instead of *txt it's txt[i]
i = 0;

Your attempt:

txt = &(*txt[0]);

Is incorrect, if you look at it. First, it takes txt[0] (which is of type char), then tries to dereference it with *, and then takes that address with the & operator (theoretically yielding the original result, though I wouldn’t want to think too hard about whether this works for dereferencing arithmetic types like char), then assigns that (arithmetic) result to a pointer. It’s like you’re saying:

const char *txt = 'a';

It doesn’t make sense. What you probably were looking for was simply

txt = &txt[0];

which, at first glance, might look like it’s setting txt to point to the first element of txt, but keep in mind that txt[0] is just *(txt + 0), or *txt. And &*txt simplifies to txt, so you’re effectively doing:

txt = txt;

Or nothing. You get the same pointer. Use one of the two methods I described.