Re: How to strip struct/class invariants?
On 7/5/15 8:15 AM, Artem Tarasov wrote: OK, so there was an old bug fixed in 2.067 (https://issues.dlang.org/show_bug.cgi?id=4421) so that now unions apparently can't contain a struct that has invariants. It kinda makes sense, although I don't see why the invariants can be simply ignored, as they don't have as much importance as destructors/postblits. More importantly, invariants are only called when you call members of a struct (or assert it). It requires actual usage of the particular union member. So already the programmer has indicated that member is valid by using it. It is NOT valid to arbitrarily pick one member of the union and call its invariant (or call all of them). I think this is an incorrect position D has taken -- Invariants should be called when calling members of a struct in a union. If an implicit choice must be made, then it should be an error. But it's easy to make the invariant call explicit. -Steve
Re: How to strip struct/class invariants?
On Sunday, 5 July 2015 at 15:39:50 UTC, Artem Tarasov wrote:
On Sunday, 5 July 2015 at 14:44:30 UTC, John Colvin wrote:
struct A
{
ubyte[B.sizeof] mem;
@property ref B b()
{
return *cast(B*)(mem.ptr);
}
mixin std.typecons.Proxy!b;
}
Thanks, I followed your suggestion and effectively rolled out
my own union implementation. Ugly but it works.
struct A
{
ubyte[maxSizeof] _data;
@property ref T _as(T)() inout { return
*cast(T*)(_data.ptr); }
alias b = _as!uint;
alias c = _as!size_t;
alias d = _as!double;
}
That works, but what I meant was, using my definition of struct
A, you can put it in a union without the compiler complaining,
and the invariants still get called. No need to write your own
union implementation.
Re: How to strip struct/class invariants?
On Sunday, 5 July 2015 at 14:44:30 UTC, John Colvin wrote:
struct A
{
ubyte[B.sizeof] mem;
@property ref B b()
{
return *cast(B*)(mem.ptr);
}
mixin std.typecons.Proxy!b;
}
Thanks, I followed your suggestion and effectively rolled out my
own union implementation. Ugly but it works.
struct A
{
ubyte[maxSizeof] _data;
@property ref T _as(T)() inout { return *cast(T*)(_data.ptr);
}
alias b = _as!uint;
alias c = _as!size_t;
alias d = _as!double;
}
Re: How to strip struct/class invariants?
On Sunday, 5 July 2015 at 12:15:32 UTC, Artem Tarasov wrote:
OK, so there was an old bug fixed in 2.067
(https://issues.dlang.org/show_bug.cgi?id=4421) so that now
unions apparently can't contain a struct that has invariants.
It kinda makes sense, although I don't see why the invariants
can be simply ignored, as they don't have as much importance as
destructors/postblits.
But more to the point. I have a struct that has an invariant,
and I wish to use it as a member of an union. With the latest
compiler, I have to somehow remove the invariant. Is there some
compile-time magic that I can use for this?
Not perfect, but I think you can do:
struct A
{
ubyte[B.sizeof] mem;
@property ref B b()
{
return *cast(B*)(mem.ptr);
}
mixin std.typecons.Proxy!b;
}
where B has an invariant. Even better, the invariant should still
get called.
How to strip struct/class invariants?
OK, so there was an old bug fixed in 2.067 (https://issues.dlang.org/show_bug.cgi?id=4421) so that now unions apparently can't contain a struct that has invariants. It kinda makes sense, although I don't see why the invariants can be simply ignored, as they don't have as much importance as destructors/postblits. But more to the point. I have a struct that has an invariant, and I wish to use it as a member of an union. With the latest compiler, I have to somehow remove the invariant. Is there some compile-time magic that I can use for this?
