On Wed, 06 Oct 2010 08:21:08 -0400, Lars T. Kyllingstad
<pub...@kyllingen.nospamnet> wrote:
On Wed, 06 Oct 2010 07:39:48 -0400, Steven Schveighoffer wrote:
Of course. If you realize that the expression [1,2,3] is not immutable,
then it makes sense.
Another example to help you think about it:
void foo(int x)
{
immutable int[3] = [1,2,x];
}
This must be run on every call of foo, because x can vary.
I don't think that is a very good reason. The compiler could detect the
special (and, might I add, common) case where an array literal whose
contents are known at compile time is assigned to an immutable variable,
and treat it as immutable even though [1,2,3] is formally of type int[].
I'm not saying it is a good explanation, but it is why the compiler does
it. Any time an array literal occurs anywhere is an allocation. I have
never seen a case where the compiler optimizes that out. Given that
context, the behavior makes sense.
I don't think any allocation should occur here, even if array literals are
not made immutable, because that allocation is going to be thrown away
immediately. I agree this should be special-cased. I believe there are
several bugs on array literals and allocations.
-Steve
