On 6/13/17 3:58 PM, ag0aep6g wrote:
On 06/13/2017 09:29 PM, Gary Willoughby wrote:
Is it possible for the `result` variable in the following code to be
returned as an immutable type if it's created by adding two immutable
types?
Qualify the return type as `inout`:
inout(Rational) opBinary(/*...*/)(/*...*/) inout {/*...*/}
(That second `inout` is the same as the first one in your code.)
Then you get a mutable/const/immutable result when you call the method
on a mutable/const/immutable instance.
It doesn't matter a lot, though. `Rational` is a value type, so you can
convert the return value between qualifiers as you want, anyway. But it
affects what you get with `auto`, of course.
Yes exactly. I prefer personally to have value types always return
mutable. You can always declare the result to be immutable or const, but
declaring mutable is not as easy. e.g.:
immutable ReallyLongValueTypeName foo1();
ReallyLongValueTypeName foo2();
version(bad)
{
auto x = foo1;
ReallyLongValueTypeName y = foo1;
}
version(good)
{
immutable x = foo2;
auto y = foo2;
}
[...]
struct Rational
{
public long numerator;
public long denominator;
public inout Rational opBinary(string op)(inout(Rational) other)
const
`inout` and `const` kinda clash here. Both apply to `this`. But
apparently, the compiler thinks `inout const` is a thing. No idea how it
behaves. I don't think it's a thing in the language. As far as I
understand, you should only put one of const/immutable/inout.
const(inout) actually *is* a thing :)
It's a type constructor that can be implicitly cast from immutable. This
has advantages in some cases.
See (horribly written) table at the bottom if the inout function section
here: http://dlang.org/spec/function.html#inout-functions
Also I talked about it last year at Dconf 2016:
http://dconf.org/2016/talks/schveighoffer.html
However, I find it very surprising that inout and const can combine the
way you have written it. Even though it does make logical sense. I'd
suspect it would be a good candidate for a "you didn't really mean that"
error from the compiler, but it might be hard to decipher from the
parsed code.
-Steve
