When applying a sort!() on a filtered range I get this compiler error:

Error: template std.algorithm.sorting.sort cannot deduce function from argument types !((a, b) => a.name < b.name)(FilterResult!(__lambda3, RangeT!(Array!(IssueType)))), candidates are:
C:\D\dmd2\windows\bin\..\..\src\phobos\std\algorithm\sorting.d(1851,1):        std.algorithm.sorting.sort(alias less = "a < b", SwapStrategy ss = SwapStrategy.unstable, Range)(Range r) if ((ss == SwapStrategy.unstable && (hasSwappableElements!Range || hasAssignableElements!Range) || ss != SwapStrategy.unstable && hasAssignableElements!Range) && isRandomAccessRange!Range && hasSlicing!Range && hasLength!Range)

But it seems the problem is with the filter!() result not being a isRandomAccessRange!Range because:
R.filter!(..).array.sort!(..) just works (by copying the filter results in a array).

Iaw is the compiler error msg wrong? Or i'm I wrong?

On Tuesday, 28 November 2017 at 13:10:15 UTC, Arjan wrote:
> Iaw is the compiler error msg wrong? Or i'm I wrong?

filter isn't random access because it doesn't even know how many elements are in there, much less where they are, until it loops through and does the comparisons to know which items pass the filter.

The .array call will buffer it and also evaluate how many items are actually still there after the filter, and sort uses that buffer too.

On 11/28/17 8:10 AM, Arjan wrote:
> When applying a sort!() on a filtered range I get this compiler error:
> 
> Error: template std.algorithm.sorting.sort cannot deduce function from argument types !((a, b) => a.name < b.name)(FilterResult!(__lambda3, RangeT!(Array!(IssueType)))), candidates are:
> C:\D\dmd2\windows\bin\..\..\src\phobos\std\algorithm\sorting.d(1851,1):        std.algorithm.sorting.sort(alias less = "a < b", SwapStrategy ss = SwapStrategy.unstable, Range)(Range r) if ((ss == SwapStrategy.unstable && (hasSwappableElements!Range || hasAssignableElements!Range) || ss != SwapStrategy.unstable && hasAssignableElements!Range) && isRandomAccessRange!Range && hasSlicing!Range && hasLength!Range)
> 
> But it seems the problem is with the filter!() result not being a isRandomAccessRange!Range because:
> R.filter!(..).array.sort!(..) just works (by copying the filter results in a array).
> 
> Iaw is the compiler error msg wrong? Or i'm I wrong?

The library is correctly telling you that your filtered range is not random access. It can't be, because it lazily applies the filter (that is, it filters on each element as you popFront them). So how can it know what the e.g. 3rd element is, if you haven't run any filters yet?

The array version works because you are applying the filter completely and storing the results elsewhere in one step.

-Steve

On Tuesday, 28 November 2017 at 13:24:09 UTC, Steven Schveighoffer wrote:
> On 11/28/17 8:10 AM, Arjan wrote:
>> [...]
>
> The library is correctly telling you that your filtered range is not random access. It can't be, because it lazily applies the filter (that is, it filters on each element as you popFront them). So how can it know what the e.g. 3rd element is, if you haven't run any filters yet?
>
> The array version works because you are applying the filter completely and storing the results elsewhere in one step.
>
> -Steve

Well I would have liked an error msg something like: isRandomAccessRange!Range for Range=.. failed! Or unable to sort!() a lazy Range.

On 11/28/17 8:57 AM, Arjan wrote:
> On Tuesday, 28 November 2017 at 13:24:09 UTC, Steven Schveighoffer wrote:
>> On 11/28/17 8:10 AM, Arjan wrote:
>>> [...]
>>
>> The library is correctly telling you that your filtered range is not random access. It can't be, because it lazily applies the filter (that is, it filters on each element as you popFront them). So how can it know what the e.g. 3rd element is, if you haven't run any filters yet?
>>
>> The array version works because you are applying the filter completely and storing the results elsewhere in one step.
>>
> 
> Well I would have liked an error msg something like: isRandomAccessRange!Range for Range=.. failed! Or unable to sort!() a lazy Range.

Sure, the error messages could be more specific about WHY it's failing. This has been discussed many times, but no action has been taken.

Would be an awesome project to add to D.

-Steve

On Tuesday, 28 November 2017 at 14:04:40 UTC, Steven Schveighoffer wrote:
> Would be an awesome project to add to D.

Oh yes, it sounds yummy..