• Eric (29/29) Apr 15 2016 1 alias J = const C;
• ag0aep6g (5/33) Apr 15 2016 Line 6 isn't accepted either. If you remove the constraint, the compiler...
• Eric (4/27) Apr 15 2016 Thanks. I see that now. Is there any way I can make the compiler
• Eric (3/34) Apr 15 2016 line 6 can be fixed like this: "const I!(J) i = a;"
• Eric (20/22) Apr 15 2016 This works:
• ag0aep6g (5/7) Apr 15 2016 A little tip: You can omit the parentheses of template instantiations
• Jonathan M Davis via Digitalmars-d-learn (8/11) Apr 15 2016 It's definitely ignored, and I'd argue that it's a bug in the compiler t...

Eric <eric x.com> writes:

   1 alias J = const C;
   2
   3 void main(string[] args)
   4 {
   5     J a = new C();
   6     I!(J) i = a;
   7 }
   8
   9 interface I(V) { }
  10
  11 class F(V) if (is(V : I!(V))) { }
  12
  13 class C : I!(J)
  14 {
  15     F!(J) m;
  16 }

The above code gives the following compile error:
Error:
template instance F!(const(C)) does not match template 
declaration F(V) if (is(V : I!V))
on line 15.

If I change line 1 to "alias J = C;" the code will compile.  The 
problem seems to be
the template constraint on line 11.  I think the constraint says, 
"If V can be automatically
converted to I!(V) then this template can be used".  However, 
line 6 does not give
an error, and it seems to be automaticaly converting J to I!(J).

-Eric

ag0aep6g <anonymous example.com> writes:

On 15.04.2016 19:13, Eric wrote:
    1 alias J = const C;
    2
    3 void main(string[] args)
    4 {
    5     J a = new C();
    6     I!(J) i = a;
    7 }
    8
    9 interface I(V) { }
   10
   11 class F(V) if (is(V : I!(V))) { }
   12
   13 class C : I!(J)
   14 {
   15     F!(J) m;
   16 }

 The above code gives the following compile error:
 Error:
 template instance F!(const(C)) does not match template declaration F(V)
 if (is(V : I!V))
 on line 15.

 If I change line 1 to "alias J = C;" the code will compile.  The problem
 seems to be
 the template constraint on line 11.  I think the constraint says, "If V
 can be automatically
 converted to I!(V) then this template can be used".  However, line 6
 does not give
 an error, and it seems to be automaticaly converting J to I!(J).

Line 6 isn't accepted either. If you remove the constraint, the compiler 
complains about it. So it's just the next error in line.

And really const C can't be converted to I!(const C) implicitly. The 
former is const, the latter is mutable => no go.

Eric <eric x.com> writes:

On Friday, 15 April 2016 at 17:43:59 UTC, ag0aep6g wrote:
 On 15.04.2016 19:13, Eric wrote:
    1 alias J = const C;
    2
    3 void main(string[] args)
    4 {
    5     J a = new C();
    6     I!(J) i = a;
    7 }
    8
    9 interface I(V) { }
   10
   11 class F(V) if (is(V : I!(V))) { }
   12
   13 class C : I!(J)
   14 {
   15     F!(J) m;
   16 }

 Line 6 isn't accepted either. If you remove the constraint, the 
 compiler complains about it. So it's just the next error in 
 line.

 And really const C can't be converted to I!(const C) 
 implicitly. The former is const, the latter is mutable => no go.


Thanks. I see that now.  Is there any way I can make the compiler
understand that the interface is const?

-Eric

Eric <eric x.com> writes:

On Friday, 15 April 2016 at 18:22:02 UTC, Eric wrote:
 On Friday, 15 April 2016 at 17:43:59 UTC, ag0aep6g wrote:
 On 15.04.2016 19:13, Eric wrote:
    1 alias J = const C;
    2
    3 void main(string[] args)
    4 {
    5     J a = new C();
    6     I!(J) i = a;
    7 }
    8
    9 interface I(V) { }
   10
   11 class F(V) if (is(V : I!(V))) { }
   12
   13 class C : I!(J)
   14 {
   15     F!(J) m;
   16 }

 Line 6 isn't accepted either. If you remove the constraint, 
 the compiler complains about it. So it's just the next error 
 in line.

 And really const C can't be converted to I!(const C) 
 implicitly. The former is const, the latter is mutable => no 
 go.


 Thanks. I see that now.  Is there any way I can make the 
 compiler
 understand that the interface is const?

 -Eric

line 6 can be fixed like this: "const I!(J) i = a;"
Now if I can just figure out how to fix line 15...

Eric <eric x.com> writes:

On Friday, 15 April 2016 at 18:28:58 UTC, Eric wrote:
 line 6 can be fixed like this: "const I!(J) i = a;"
 Now if I can just figure out how to fix line 15...

This works:

   1 alias J = const C;
   2
   3 void main(string[] args)
   4 {
   5     J a = new C();
   6     const (I!(J)) i = a;
   7 }
   8
   9 interface I(V) { }
  10
  11 class F(V) if (is(V : const(I!(V)))) { }
  12
  13 class C : const (I!(J))
  14 {
  15    F!(J) m;
  16 }
  17

-Eric

ag0aep6g <anonymous example.com> writes:

On 15.04.2016 20:55, Eric wrote:
   13 class C : const (I!(J))

I think this const is ignored by the compiler.

   15    F!(J) m;

A little tip: You can omit the parentheses of template instantiations 
when the argument is a single identifier. That is, `F!(J)` can be 
written as `F!J`.

Jonathan M Davis via Digitalmars-d-learn writes:

On Saturday, April 16, 2016 00:28:46 ag0aep6g via Digitalmars-d-learn wrote:
 On 15.04.2016 20:55, Eric wrote:
   13 class C : const (I!(J))

 I think this const is ignored by the compiler.

It's definitely ignored, and I'd argue that it's a bug in the compiler that
it's even accepted. There's no such thing as a const or immutable class -
just a const or immutable instance of a class. Member functions of a class
can be const or immutable but not the class itself.  Doing so is just
equivalent to marking every function in the class as const or immutable.
Deriving from a const or immutable class is nonsense.

- Jonathan M Davis