On Fri, 05 Nov 2010 05:00:10 -0400, Lars T. Kyllingstad
<[email protected]> wrote:
On Fri, 05 Nov 2010 00:00:53 -0400, ponce wrote:
4.
inout(T) myFunction(inout(T) x)
{
inout(T) a;
}
Can we declare a inout(T) variable in an inout(T) function ?
Currently you can, but I don't know whether that's intended or not.
(AFAIK, inout is only halfway implemented in the compiler.)
Yes, it's intended. You can only declare inout variables on the stack,
you cannot declare them as members of a struct/class.
However, note that you can only assign inout(T) to inout(T), it doesn't
accept any other types.
5.
T func(T x)
{
// blah
}
const(T) func(const(T) x)
{
// blah
}
immutable(T) func(immutable(T) x)
{
// blah
}
inout(T) func(inout(T) x)
{
// blah
}
How is the overloading resolved when calling func(x) where x is of type
T/const(T)/immutable(T)/inout(T) ?
The closest match is always chosen, and inout(T) is last resort.
When x is mutable T, the compiler looks for, in this order,
1. func(T)
2. func(const T)
3. func(inout T)
When x is const(T), the compiler looks for
1. func(const T)
2. func(inout T)
When x is immutable(T), the compiler looks for
1. func(immutable T)
2. func(const T)
3. func(inout T)
This is correct. The idea is to handle the common case, and then if
specializations are desired, they can be implemented separately.
-Steve
