protocol for using InputRanges

[Oscar Martín" <omarmed0000 at hotmail.com>]

On Friday, 28 March 2014 at 23:14:56 UTC, Tobias Müller wrote:
> On Thursday, 27 March 2014 at 20:49:16 UTC, Walter Bright wrote:
>> On 3/27/2014 12:21 PM, Rainer Schuetze wrote:
>>> This loop is intuitive. Not being allowed to call empty or 
>>> front multiple times
>>> or not at all is unintuitive. They should not be named as if 
>>> they are properties
>>> then.
>>
>> I can concede that. But I can't concede being able to call 
>> front without first calling empty, or calling popFront without 
>> calling empty and front, or requiring 'pump priming' in the 
>> constructor.
>
> Disclaimer: I'm a C++ programmer just lurking here, I've never
> actually used D.
>
> I find it very counter-intuitive that 'empty' is required before
> front or popFront.
> Since 'pump priming' in the constructor isn't wanted either, i'd
> suggest the following protocol:
>
> while (popFront())
> {
>     front;
> }
>
> popFront is then required to return !empty.
> 'empty' as a separate property getter can stay but is not
> required for the protocol.
>
> This way, it's clear that the work to fetch the next element is
> always done in popFront.
>
> Generally I find dependecies between functions problematic that
> require a specific call sequence. If they can be removed, they
> should.
>
> With my proposed solution, there's still one minor dependency,
> namely that front is not valid before the first popFront. This
> could be solved by again combining the two, as proposed by
> someone else in this thread.
>
> Tobi

First sorry for my english.

I agree.

All work is done in "bool popFront()" (well, this name may no 
longer be the most appropriate).
In this function the first / next item is obtained and caches. 
Returns true if any element, false otherwise.
"front" return de cached element as now, as many times as desired 
and without side effects.
"empty" function is no longer necessary, but it might be useful 
to keep changing the return type for example to int:

     1 - Sure there is more elements
     0 - Sure there is NO more elements
    -1 - I don't known. Try popFront to know if more element

Of course this function as "front", without side effects