On 16/07/10 02:08, Jonathan M Davis wrote:
On Thursday, July 15, 2010 17:40:26 Gareth Charnock wrote:
So having got a collectors' edition TDPL, I though I'd have a try at
writing some concurrent code. The idea was a worker thread(s) would do
some work and write the results to some immutable objects. These would
get passed to an indexer thread that would do neat stuff like indexing
the references to the objects in things like hash tables and such,
performing any reduce type operations along the way. These views would
then be passed back to workers in due course. Unfortunately I hit a
snag. It appears you can't do this:
class A {
...
}
immutable A a1 = new immutable(A); //pointer to a real, immutable object
immutable A a2; //pointer to null
a2 = a1; //error, guess I doomed a2 to a life of being useless
I thought that as such assignments don't change the underlying objects
in memory, they would be fine. Compare to:
immutable(char)[] s1 = "hello world"; //fat pointer to a section of
immutable memory
immutable(char)[] s2; //fat pointer to null
s2 = s1; //fine
The only solutions I can think of:
1) Use a pointer to a class. Works, but that just seems very unsafe and
just plain un-D-ish. We also have to dereference two pointers.
2) Work entirely with immutable(A)[], which is not quite as crazy as it
seems as we are copying about arrays of pointers rather than arrays of
A. It's still quite crazy.
3) Write some sort of wrapper struct to hide the pointer.
4) Perhaps ref works like C++ int&?
ref immutable(A) ref_a //error
Nope.
So is this intended behavior? Am I missing something obvious?
First off, you're using using a reference, not a pointer. They're similar but
quite different. If you were using a pointer, you could do
immutable (A)* a1;
and the object would be immutable while the pointer would be mutable. It's a
mutable pointer to an immutable object of type A. The problem with a reference
is that it has no syntactic way to differentiate between what's doing the
refering and what's being referred. Because immutable is transitive,
immutable A a1;
is an immutable reference to an immutable object of type A. As soon as you try
and make a reference immutable, what it refers to is immutable as well. There is
no syntactic way to fix the problem. The solution is Rebindable!(T) in
std.typecons.
Rebindable!(T) is a wrapper struct. It allows you to have const and immutable
references and still change what they refer to. Take this example from the docs:
// Rebindable references to const and immutable objects
void bar()
{
const w1 = new Widget, w2 = new Widget;
w1.foo();
// w1 = w2 would not work; can't rebind const object
auto r = Rebindable!(const Widget)(w1);
// invoke method as if r were a Widget object
r.foo();
// rebind r to refer to another object
r = w2;
}
By wrapping the const (or immutable) Widget, you can "rebind" what is "bound" by
Rebindable!(T) and effectively get a mutable reference to a const or immutable
object. It's not terribly pretty, but apparently no one could come up with a
satistfactory way of doing it in the language itself given the syntax for
references. So, Rebindable!(T) is the solution.
- Jonathan M Davis
Thanks, that should work just fine. Looking at the phobos source for
Rebindable, I was nowhere near close to reinventing that wheel.
