On Fri, 16 Jul 2010 12:12:38 +0200, Rory McGuire wrote:
> On Fri, 16 Jul 2010 11:58:57 +0200, Lars T. Kyllingstad
> <pub...@kyllingen.nospamnet> wrote:
>
>> On Fri, 16 Jul 2010 11:46:48 +0200, Rory McGuire wrote:
>>
>>> import std.stdio;
>>>
>>> struct State {
>>> string s; string getString() { return s; } static State
>>> opCall(string s) {
>>> State ret;
>>> ret.s = s;
>>> return ret;
>>> }
>>> }
>>>
>>> void main() {
>>> auto s = State("adf");
>>> pragma(msg, s.getString());
>>> }
>>>
>>> dmd Output: (line 14 is the pragma statement)
>>>
>>> struct.d(14): Error: variable s cannot be read at compile time
>>> struct.d(14): Error: cannot evaluate s.getString() at compile time
>>> s.getString()
>>
>> It's not working because s isn't a compile-time quantity. Try:
>>
>> enum s = State("adf");
>>
>> -Lars
>
> Awesome thanks, worked.
>
> So is the difference that "auto s" is a Struct which can change whereas
> "enum s" is a constant?
> If it is a constant its just "s" that is constant right?
>
> Thanks Lars
Yes. Writing "auto s = State("adf");" is equivalent to writing
State s = State("adf");
and since s can change at runtime, it would be meaningless to say that it
has a value at compile time. However, its *initial value*, the struct
literal State("adf"), is known at compile time -- otherwise it would be
impossible to assign it to an enum.
And actually, an enum is not only a constant, it is a manifest constant.
When you declare "enum x = someFixedValue", the compiler will just
replace all later uses of x with someFixedValue. It's basically the same
as using the literal value in those places.
This has the consequence that you can't take the address of an enum, nor
pass an enum by reference. Doing
enum i = 3;
int* p = &i;
is equivalent to
int* p = &3;
which doesn't make sense.
-Lars
