Re: When to use "call by reference"?
Jacky wrote:
if I have following statements...
int *foobar;
class foobar2
{
};
foobar2 foobarx = new foobar2();
This already won't compile.
would I choose this one
int foo(int *foobar); //copy foobar pointer to foo
or
this
int foo(int& *foobar); // no copy but pass foobar as an altered
pointer...
which of these is more efficient
Impossible to tell. Please, for your own sake, refrain from thinking in
terms of performance for now, i.e. until you have the C++ basics down. The
two functions simply do different things, at least they implicitly say so.
The first one says "this function takes a pointer, which can be null, to an
int that is modified inside the function".
The second one says "this function takes a reference to a pointer. Inside
the function both the pointer as well what it points to will be modified".
is this legal?
Yes, but it is probably bad design, too. Please write down in plain English
what the intention of a function is. Without that, every discussion how it
should do its job is pointless.
int foo(const int& *foobar); //no no?
I think that one is covered in the C++ FAQ, though it might refer to
a 'reference to a reference' instead of a 'pointer to a reference'.
int foo(foobar2 *foobarx); // copied class pointer
Same as 'int foo(int* p);'.
int foo(foobar2 &foobarx); // non-copied changable pointer
"This function takes a reference (i.e. something that can not be null) to a
foobar2 instance and modifies it."
Uli
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