Call a function that accepts char * as const char *&.

[Sajas K K]

Can i have something like a reference to pointer that can change the pointer but not the value in the pointed location.

void fail(const char *& i)
{
    i++;
}


int main()
{
    char s[]="something";
    char *j=s;
    fail(j);
}

[Salem]

I just changed it to
const char *j=s;

and it does what I expected it to.

[Elkvis]

the code you posted will increment the pointer itself. following the call to fail() in main(), j will now point to the 'o' in "something." the traditional (C) way of doing this was to pass a pointer to the pointer (char**), but C++ allows a reference to a pointer (but not the other way around), so you can modify the pointer itself without needing to dereference anything.

[Elysia]

But why do you want to do it?

[Elkvis]

But why do you want to do it?

indeed. it would be simpler to do this:

char* fail(char* ptr)
{
  return ++ptr;
}

[Elysia]

That was not my point. Why the need to pass a char pointer in C++?