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December 07, 2007
Invariants broken with out cast!!!
An invariant(int) can be assigned to an invariant(int)*, this wasn't
possible in 2.007. Just compile the attach test case.
December 07, 2007
Re: Invariants broken with out cast!!!
None of that matters any more. It's all change again, and this time I
do believe we've got it right.

We can worry about bugs in that next generation behaviour after it's
implemented.
December 07, 2007
Re: Invariants broken with out cast!!!
On Thu, 06 Dec 2007 20:43:12 -0800, Robert DaSilva wrote:

> An invariant(int) can be assigned to an invariant(int)*, this wasn't
> possible in 2.007. Just compile the attach test case. import std.stdio;
> 
> void main()
> {
>     invariant(int) a;
>     invariant(int)* b = &a; // error in 2.007, ok in 2.008 writeln(*b);
>     a = 1;
> //    *b = 2; // error
>     writeln(*b);
> }

Even if it's broken, the behaviour seems logical to me.
The types aren't different, so why would you need a cast?

Just my 2 cents.
December 07, 2007
Re: Invariants broken with out cast!!!
On Thu, 06 Dec 2007 20:43:12 -0800, Robert DaSilva wrote:

> An invariant(int) can be assigned to an invariant(int)*, this wasn't
> possible in 2.007. Just compile the attach test case. import std.stdio;
> 
> void main()
> {
>     invariant(int) a;
>     invariant(int)* b = &a; // error in 2.007, ok in 2.008 writeln(*b);
>     a = 1;
> //    *b = 2; // error
>     writeln(*b);
> }

Then again, I'd expect *b to be mutable if a is mutable.
Someone care to explain what the correct behaviour should be and why?
December 07, 2007
Re: Invariants broken with out cast!!!
On Fri, 07 Dec 2007 08:17:34 +0000, Denton Cockburn wrote:

> On Thu, 06 Dec 2007 20:43:12 -0800, Robert DaSilva wrote:
> 
>> An invariant(int) can be assigned to an invariant(int)*, this wasn't
>> possible in 2.007. Just compile the attach test case. import std.stdio;
>> 
>> void main()
>> {
>>     invariant(int) a;
>>     invariant(int)* b = &a; // error in 2.007, ok in 2.008 writeln(*b);
>>     a = 1;
>> //    *b = 2; // error
>>     writeln(*b);
>> }
> 
> Then again, I'd expect *b to be mutable if a is mutable. Someone care to
> explain what the correct behaviour should be and why?

Nevermind, I think I got it.

a is a mutable ref to an invariant int.
b is a pointer to an invariant int
thus: *b is an invariant int that cannot be reassigned
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