On Mon, 05 Mar 2012 11:17:41 -0500, Stewart Gordon <smjg_1...@yahoo.com> wrote:

On 05/03/2012 13:49, Steven Schveighoffer wrote:
On Sat, 25 Feb 2012 09:02:47 -0500, Timon Gehr <timon.g...@gmx.ch> wrote:
<snip>
inout means 'some qualifier but we don't know which one'. The call site on the other hand knows which qualifier it is. If the call is made to work there is no 'making it mutable', because it is mutable all the time. The inout function cannot change the data
because it cannot know what the constancy is.

What you would propose is that a delegate which takes a const/mutable/immutable implicitly translates to one which takes an inout. I agree it can be made to work, but it does not
fit into the current definition of inout.

I'm not sure if you understand correctly. Is isn't a matter of implicit conversion of delegates from one type to another, but a matter of (in this case) opApply accepting a delegate with a parameter whose constancy matches that of this.

I understand the problem and the proposed solution quite well.

It would be ideal for inout to solve this, but it's not chartered in such a way to do so.

If I'm not mistaken, inout isn't chartered to do anything particular when it occurs in a function signature nested within another. The spec just leaves it ambiguous.

No, inout is not modifiable during the function execution. Given the transitivity of const and immutable, inout itself must also be transitive.

Think about it this way:

inout(int)* foo(inout(int)* x)
{
   // begin here, at this point x is not modifiable
   // *x = 5; // not allowed!
   return x;
// end here, at this point inout reverts back to it's original constancy (including return value).
}

The original definition I used for inout was "scoped const", meaning it's const within the scope and reverts back after the scope. I did not plan for having the constancy "temporarily" revert back to the original constancy. What you wish for is this to take a delegate which matches the constancy of the x parameter:

inout(int)* foo(inout(int) *x, void delegate(inout(int)* p) dg)
{
   dg(x); // valid, since the type matches
return x; // oops, now x could have changed! Even through it's an inout reference!
}


It's currently transitive, and this would break transitivity. If we want to look at fundamentally redefining inout so that it can break transitivity, then we can look at that. But I don't think this is a simple "add-on" to the current functionality.
<snip>

Can you give an example of how it breaks transitivity?

Using the above foo function:

void main()
{
   void bar(int *x) {*x = 5;}
   int i = 2;
foo(&i, &bar); // this sets i to 5 via the inout reference we pass into it.
}

In other words, inout is transitively not constant during the function call.

I'm not saying we cannot bend the rules to allow for this, because clearly it doesn't violate the "true" constancy type of the data passed in, but I don't think it's a straightforward change conceptually. I'm not a compiler writer, so I don't know how this works in their minds, I'd like to have their input.

-Steve

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