On 03/07/2012 04:41 PM, Steven Schveighoffer wrote:
On Tue, 06 Mar 2012 05:11:34 -0500, Timon Gehr <timon.g...@gmx.ch> wrote:
On 03/06/2012 12:27 AM, Steven Schveighoffer wrote:
...
There are two parts to inout, one is that it can be one function called
3 different ways, the other is that you know it's constant during
function execution. Some people like that second part, even if it
doesn't fully guarantee everything. I.e. there's a reason people use
const in C++ besides it being trendy.
By passing a delegate that changes an inout-matched argument it is
made explicit that inout data may change. Technically, none of the
existing guarantees are lost, we just add some expressiveness to the
type system.
Yes, I understand that it works. I know that it doesn't violate
technically any const guarantees.
Yes, that is not what I was after:
void foo(inout(int)*x, void delegate(inout(int)*) dg)
// both inout's denote the same wildcard
{ dg(x);
}
void main(){
int x;
foo(&x, (p){*p=2;}); // this currently cannot be valid code
}
The function call is explicit about that may change the inout parameter.
There are no guarantees lost. It is the call site who changes the
parameter. Code that currently has the guarantee that the parameter
won't change will keep it, because there is no valid code that could
silently change semantics under the change and does not use some fancy
introspection.
