the naive version would look like

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

On Wednesday, 15 June 2022 at 13:52:24 UTC, Christian Köstlin wrote:

Because you have a template function, most attributes will be inferred based on the Range type. const is not inferred, but popFront mutates so it doesn't apply in your case. @property is not inferred either, but the spec says "Using property functions is not recommended until the definition is more certain and implementation more mature."

On Wednesday, 15 June 2022 at 13:52:24 UTC, Christian Köstlin wrote:

auto next(Range)(Range r) {
    r.popFront;
    return r.front;
}

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.

On Wednesday, 15 June 2022 at 17:30:31 UTC, JG wrote:

auto next(Range)(Range r) {
    r.popFront;
    return r.front;
}

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.

auto next(Range)(Range r) {
  auto ret = r.save;
  ret.popFront;
  return ret;
}

On 2022-06-15 19:36, JG wrote:

auto next(Range)(Range r) {
    r.popFront;
    return r.front;
}

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

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

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 ...