I do not understand copy constructors

[Steven Schveighoffer]

On 8/12/21 10:08 AM, Learner wrote:
> On Thursday, 12 August 2021 at 13:56:17 UTC, Paul Backus wrote:
>> On Thursday, 12 August 2021 at 12:10:49 UTC, Learner wrote:
>>>
>>> That worked fine, but the codebase is @safe:
>>>
>>> ```d
>>> cast from `int[]` to `inout(int[])` not allowed in safe code
>>> ```
>>>
>>> So copy constructors force me to introduce trusted methods, while 
>>> that was not necessary with postblits?
>>
>> A postblit would simply ignore the type qualifier--which can lead to 
>> undefined behavior. (Scroll down to the paragraph that begins "An 
>> unqualified postblit..." under ["Struct Postblits"][1] in the spec.) 
>> The copy constructor merely forces you to be honest about the safety 
>> of your code.
>>
>> In your case, I would recommend encapsulating the unsafe cast in a 
>> function like the following:
>>
>> ```d
>> T[] dupWithQualifiers(T[] array)
>> {
>>     auto copy = array.dup;
>>     return (() @trusted => cast(T[]) copy)();
>> }
>> ```
>>
>> You can then use this function in place of `dup` in your copy 
>> constructor.
>>
>> [1]: https://dlang.org/spec/struct.html#struct-postblit
> 
> Thank you, now everything is more clear.
> 
> A last question, if you do not mind, just to better understand inout. It 
> seems a shortcut to avoid repeating the same function body for mutable, 
> const, and immutable. Why the following code is not equal to the single 
> inout constructor?
> 
>      struct A {
>          int[] data;
> 
>          //this(ref return scope inout A rhs) inout { /*body*/ }
> 
>          this(ref return scope           Timestamp rhs) { /*body*/ }
>          this(ref return scope const     Timestamp rhs) const { /*body*/ }
>          this(ref return scope immutable Timestamp rhs) immutable { 
> /*body*/ }
>      }
>      Error: Generating an `inout` copy constructor for `struct B` 
> failed, therefore instances of it are uncopyable
> 
> Inout is compatible only with inout, and not with the unrolled code it 
> implies?

inout is not like a template. It's a separate qualifier that generates 
only one function (not 3 unrolled ones).

It's sort of viral like const is viral -- all underlying pieces have to 
support inout in order for you to write inout functions.

-Steve