On Tue, 07 Sep 2010 09:28:18 -0400, Pelle <pelle.mans...@gmail.com> wrote:
On 09/07/2010 03:15 PM, Steven Schveighoffer wrote:
On Tue, 07 Sep 2010 08:56:15 -0400, Jacob Carlborg <d...@me.com> wrote:
On 2010-09-07 14:49, Steven Schveighoffer wrote:
On Sun, 05 Sep 2010 09:40:59 -0400, BLS <windev...@hotmail.de> wrote:
On 05/09/2010 02:16, Jonathan M Davis wrote:
void foo(T)(T[] collection, T elem)
{
// Blah, whatever
}
I am curious, how this will look and feel once inout is working ?
inout void foo(T)(inout(T)[] collection, inout T elem)
{
// Blah, whatever}
}
inout void doesn't make any sense. You can't have a const void or
immutable void.
Now, if foo is a member function, then inout applies to the "this"
pointer, but even then, you need a return type other than void for
inout
to be used.
-Steve
inout is only used when you want to return the same constness
(mutable, const, immutable) as you passed in to the function. If you
don't want that, or don't want to return anything then const(T)[] is
what you want. It will accept mutable, const and immutable.
Yes, exactly. This is why inout functions cannot return void.
-Steve
Hmm.
class C {
void foo(void delegate(inout(C)) f) inout {
f(this);
}
}
Am I missing something?
Yes, a valid return. Your function should be:
void foo(void delegate(const(C) f) const
It helps to understand that inout/const/immutable has NOTHING to do with
code generation, it only has to do with limiting what compiles. For this
reason, an inout function is compiled once, and works on all three
constancies (4 if you have a nested inout function). For the entire
function any inout variable is treated as a non-changeable value, just
like const. Then when you return, it's converted at the call site back to
the constancy with which it was called. If the return value is void, then
there's nothing to convert, and no reason to use inout over const.
I'll repeat -- there is no benefit to inout if you are not returning
anything.
-Steve
