On 2012-10-04 16:18, Ali Çehreli wrote:
inout is like a template on 'mutable', const, and immutable; but it need
not be applied to templates. Here is a simple example that transfers the
mutability to the return type:
I do think I understand how "inout" works, or at least I thought. I
don't understand how the delegate affects anything. If I remove the
delegate to this code it compiles:
void foo (inout int[] arr)
{
auto a = arr[0];
}
My actual problem is that I want to be able to pass both mutable, const
and immutable to a function. In that function I want the "mutability" to
persist.
I first used const for my function and that works, kind of. The problem
if I do like this:
void foo (const int[] arr)
{
auto s = typeid(typeof(arr)).toString;
// here "s" will be something like "const(const(int)[])"
}
I doesn't matter if I pass immutable or mutable, "s" will always be
"const(const(int)[]". So I thought, "hey, that's what inout is for, I
use that", then everything blows up.
I'm using "typeid(typeof(arr))" in my serialization library to get a
string representation of a type. I would prefer that the mutability was
correct.
--
/Jacob Carlborg
