Whats the proper way to write a Range next function

[Christian Köstlin]

On 2022-06-15 19:36, JG wrote:
> On Wednesday, 15 June 2022 at 17:30:31 UTC, JG wrote:
>> On Wednesday, 15 June 2022 at 13:52:24 UTC, Christian Köstlin wrote:
>>> the naive version would look like
>>>
>>> ```d
>>> auto next(Range)(Range r) {
>>>     r.popFront;
>>>     return r.front;
>>> }
>>> ```
>>>
>>> But looking at a mature library e.g. 
>>> https://github.com/submada/btl/blob/9cc599fd8495215d346ccd62d6e9f1f7ac140937/source/btl/vector/package.d#L229 
>>> is looks like there should be tons of annotations/attributes on it.
>>>
>>> Kind regards,
>>> Christian
>>
>> I not sure of your use case. But you need to check if the range is 
>> empty before and after calling popFront (and decide what to if it is). 
>> Also unless you want the original range passed in mutated you should 
>> call save (assuming you have a forward range) before calling your next 
>> or you need to modify next so it calls save.
> 
> This is what I would write for next (which should only be called after
> checking if the range is empty).
> 
> It produces a range whose front points at the next element.
> ```d
> auto next(Range)(Range r) {
>    auto ret = r.save;
>    ret.popFront;
>    return ret;
> }
> ```
Thanks for the explanation!
I actually want to mutate the original range and I know, that there are 
at least 2 elements in my range.

e.g. the program
```d
import std.stdio;
import std.range;

auto next(Range)(Range r) {
     r.popFront;
     return r.front;
}
int main(string[] args) {
     auto i = iota(10);
     writeln(i.next);
     writeln(i.next);
     return 0;
}
```
prints:
```
1
1
```

whereas
```d
import std.stdio;
import std.range;

auto next(Range)(ref Range r) {
     r.popFront;
     return r.front;
}
int main(string[] args) {
     auto i = iota(10);
     writeln(i.next);
     writeln(i.next);
     return 0;
}
```
prints
```
1
2
```
which is what I would want. So it depends if the Range is a value or 
reference type ...