Ranges and backward iteration

[Andrew]

On Sunday, 29 July 2012 at 23:42:09 UTC, Timon Gehr wrote:
> On 07/30/2012 01:26 AM, Andrew wrote:
>> I have a use case where I would like to be able to pass both a
>> forward and backward iteration of an array to a function:
>>
>> void foo(InputR, OutputR)(InputR i, OutputR j)
>> if (isInputRange!InputR && isOutputRange!(OutputR, InputR))
>> {
>> ...
>> }
>>
>> main()
>> {
>> //forward:
>> foo(a[1..n], b[1..n]);
>> //backward:
>> foo(retro(a[n..$]), retro(b[n..$]));
>>
>> //do stuff with b as it has been constructed now...
>> }
>>
>> This doesn't work
>
> Works for me.
>
> This is how I test:
>
> void foo(InputR, OutputR)(InputR i, OutputR j) if 
> (isInputRange!InputR && isOutputRange!(OutputR, InputR)){
>     j.put(i);
> }
>
> void main() {
>     int[] a=[2,0,0,3], b=[1,2,3,4];
>     int n=2;
>     foo(a[0..n], b[0..n]);
>     assert(b==[2,0,3,4]);
>     foo(retro(a[n..$]), retro(b[n..$]));
>     assert(b==[2,0,0,3]);
> }
>
>> (retro doesn't appear to return a Range, but
>> rather an object of type "Result", which I don't understand)
>
> 'Result' implements the range interface and is a local struct 
> of the 'retro' function.
>
>>, but the intent should be clear.
>>
>> How do I do this? What am I missing? There doesn't seem to be a
>> "backward iterator" equivalent in std.range.
>
> retro should do the job.

Awesome, thanks... In my code, I had a "ref" for the second
argument, which apparently made the type signature not work.