How to break const
Jonathan M Davis
jmdavisProg at gmx.com
Mon Jun 18 00:58:29 PDT 2012
On Monday, June 18, 2012 08:19:55 Mehrdad wrote:
> On Monday, 18 June 2012 at 06:14:22 UTC, Matthias Walter wrote:
> > Its not, that a const method cannot modify an object, it just
> > ensures that the const method cannot modify the object *by
> > using the this-pointer*.
>
> I see...
>
>
> So that means you /can't/ tell something just by looking at a
> part of the code, right?
>
> (Just mentioning this since this idea seemed to be emphasized a
> lot by D.)
You can if the function is pure as well, because then that delegate would not
be legal. For instance, this code
import std.stdio;
struct Const
{
this(void delegate() pure increment)
{ this.increment = increment; }
int a;
void delegate() pure increment;
void oops() const pure { this.increment(); }
}
void main()
{
Const c;
c = Const({ c.a++; });
writeln(c.a);
c.oops();
writeln(c.a);
}
fails to compile, giving this error:
q.d(15): Error: constructor q.Const.this (void delegate() pure increment) is
not callable using argument types (void delegate() nothrow @safe)
All const guarantees is that the object isn't altered through the const
reference/pointer (which in the case of a const function is this). That's
powerful, but it needs pure as well to really be able to just glance at it and
know that it's not altering your object.
If you want an extreme exampl. you could create an object whose entire state
was held in a global variable, then the fact that the this pointer was const
wouldn't mean much. But if the member functions were pure, then you couldn't
access that global variable, and so that externalization of the state wouldn't
work, and you'd be guaranteed that the const function didn't alter your object
(as long as none of the arguments to that const function held a reference or
pointer to that object anyway (though that's a fairly abnormal thing to do) -
only strong purity absolutely guarantees that your object isn't being
altered).
- Jonathan M Davis
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