http://d.puremagic.com/issues/show_bug.cgi?id=7355
--- Comment #5 from Steven Schveighoffer <schvei...@yahoo.com> 2012-01-26
09:36:01 PST ---
(In reply to comment #3)
> The typeof resolves to error because inout resolves to immutable.
As I said, it should fail to match or match mutable and fail to call. I'm not
sure which is correct, but I feel either way that the assert should fail. If
it's resolving to immutable, I think it's a bug, not because it's not passing,
but because it's failing for the wrong reason.
I think your expectations would be a violation of const. Let's assume inout
did resolve to const for foo, and the function could be called:
immutable int x = 5;
immutable(int)* xp = &x;
immutable(int)** xpp = &xp;
const(int *)* y = foo(xpp);
int z = 2;
*y = &z; // this should pass, since I can assign int* to const(int*).
assert(*xp == 2);
z = 3;
assert(*xp == 3); // oops! changed data perceived as immutable!
Is there an error in my example? I think it comes down to this:
immutable(int *)* foo(immutable(int *)* x) // inout == immutable
const(int *)* foo( const(int *)* x) // inout == const
int ** foo( int ** x) // inout == mutable
none of these can be called with immutable(int)** because there is no implicit
cast to the parameter. I don't think const(immutable(int)*)* reduces to
const(int *)*.
This does take some mental effort, so I may have made a mistake :) I hate
double pointers...
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