On 1/29/18 8:20 PM, Jonathan M Davis wrote:
On Tuesday, January 30, 2018 01:05:54 Drone1h via Digitalmars-d-learn wrote:
Hello all,
I am trying to implement a ("struct template" ? what is the
correct word ?) range that just forwards its primitives ("empty",
"front", "popFront") to another range, possibly with some very
limited filtering/alteration, as std.range.Take (just to learn).
Initially, the "front" member function (property) used to be
declared "const", but that was not accepted when the underlying
range (denoted "R" in the code below) was std.stdio.File.ByChunk
("Error: mutable method std.stdio.File.ByChunk.front is not
callable using a const object").
Is there any value in having the "front" range primitive declared
to be a "const" member function ?
And if so, is the following implementation okay ? Could it be
further simplified ?
struct Taker (R)
{
private R _r;
...
static if (functionAttributes ! (R.front) &
FunctionAttribute.const_)
public @property auto front () const { return
_r.front; }
else
public @property auto front () { return
_r.front; }
...
}
Thank you respectfully !
Drone1h
If you want to put an attribute on it, inout is better, because then it will
work with any constness, but in general, I'd suggest just avoiding the use
of const or immutable altogether when dealing with ranges. front can return
a const element, and that will happen if you use auto and whatever you're
wrapping is const, but const ranges are utterly useless, because they can't
be mutated and thus can't be iterated. As such, almost no code is ever going
to have a range that is anything but mutable, which means that having front
be anything but mutable is generally pointless.
Not necessarily. A main reason for const is to advertise "I'm not going
to change your mutable data" on a function. So reading that front is
const (or inout more appropriately) can assure you front makes this
guarantee.
Yes, it also allows you to call on an immutable or const range, both of
which are for the most part useless.
So I would say const ranges are useless, but const members of ranges
provide some value.
That being said, const is viral, as is inout. So unfortunately if you
*don't* mark your functions const or inout, then wrappers need to take
this into account.
-Steve
