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July 29, 2012
Ranges and backward iteration
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 (retro doesn't appear to return a Range, but
rather an object of type "Result", which I don't understand), 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.
July 29, 2012
Re: Ranges and backward iteration
On Sunday, 29 July 2012 at 23:26:09 UTC, 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 (retro doesn't appear to return a Range, but
> rather an object of type "Result", which I don't understand), 
> 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.



 isBidirectionalRange perhaps?

[quote]
Returns true if R is a bidirectional range. A bidirectional range 
is a forward range that also offers the primitives back and 
popBack. The following code should compile for any bidirectional 
range.
[/quote]
July 29, 2012
Re: Ranges and backward iteration
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.
July 29, 2012
Re: Ranges and backward iteration
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.
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