inout and function/delegate parameters

Timon Gehr timon.gehr at
Sat Feb 25 06:02:47 PST 2012

On 02/24/2012 05:26 PM, Steven Schveighoffer wrote:
> On Sun, 19 Feb 2012 09:27:42 -0500, Stewart Gordon <smjg_1998 at>
> wrote:
>> At the moment, if a function has an inout parameter, it must have an
>> inout return type.
>> But this prevents doing stuff like
>> void test(ref inout(int)[] x, inout(int)[] y) {
>> x = y;
>> }
> This is a legitimate request, I think there is an effort underway by
> myself Timon and Kenji to make this work. (well mostly Kenji and Timon)
>> or passing the constancy through to a delegate instead of a return value.
>> A typical use case of the latter is to define an opApply that works
>> regardless of the constancy of this and allows the delegate to modify
>> the iterated-through objects _if_ this is mutable.
>> int opApply(int delegate(ref inout(T)) dg) inout;
> What you ask isn't possible given the current design of inout. During
> inout function execution, inout is a special form of const, even if the
> object on which opApply is being called is mutable.

The call site has enough information to type check the call given that 
there is some syntax to tie the two inout qualifiers together. inout 
shouldn't imply const, we already have const for that.

>> But then I realised a potential ambiguity:
>> (a) the constancy is passed through to the delegate
>> (b) the delegate has an inout parameter in its own right
>> If we go by interpretation (b), then each signature contains only one
>> inout, so even if we relaxed the rules to allow this it would just be
>> equivalent to
>> int opApply(int delegate(ref const(T)) dg) const;
> Yes, this is what I think it should be equivalent to. As I said, inside
> opApply, inout is like const, and is transitive. So you cannot
> "temporarily" make it mutable.

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.

>> however, this won't always be true in the general case.
>> The essence of functions with inout parameters is that they have a
>> hidden constancy parameter. This is essentially a template parameter,
>> except that only one instance of the function is generated, rather
>> like Java generics. If we made this parameter explicit in the code, we
>> could distinguish the two meanings:
> No the constancy is not a parameter (not even a hidden one). The magic
> of inout happens at the call, not inside the function.

The same is (mostly) true for Java generics.

> Inside, it's just another type of const.
> However, there is an entire part of delegates that is yet untapped --
> implicit conversion of delegates. For example, int delegate(ref const
> int x) could be implicitly converted to int delegate(ref int x). Maybe
> there is something there that can be used to solve this problem. Maybe
> there is a rule we could apply when calling an inout-enabled function
> that allows implicit conversion of a delegate to an inout -flavored
> version of the delegate (as long as the normal delegate matches the
> decided inout constancy factor). But that violates transitivity. I'm not
> sure Walter would go for this.

Unfortunately this violates type safety. Also see:

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