pure vs std.range.retro

[John Carter]

Thanks for all your help understanding the algorithm chaining problem...

Now I have another gotcha.

This code compiles and runs OK if I remove the keyword "pure" or if I
remove the ".retro"

As I understand it retro should affect the "pure"ness of the the function.

If I put the keyword in, it fails to instantiate the template...
import std.algorithm;
import std.array;
import std.stdio;
import std.range;

pure auto huh( in uint[] digitArray, in uint value)
{
   return find!( (a,b) => (a > b))( digitArray.retro, value);
}

unittest {
   auto test = [11u, 1u, 1u, 10u];
   auto s = huh( test, 10u);
   writeln( s);
}

dmd -unittest -main  doesnt_work.d && ./doesnt_work
/usr/include/dmd/phobos/std/algorithm.d(3508): Error: pure function
'doesnt_work.huh' cannot call impure function
'std.range.retro!(const(uint)[]).retro.Result.popFront'
/usr/include/dmd/phobos/std/algorithm.d(3510): Error: pure function
'doesnt_work.huh' cannot call impure function
'std.range.retro!(const(uint)[]).retro.Result.front'
doesnt_work.d(8): Error: template instance doesnt_work.huh.find!(__lambda3,
Result, const(uint)) error instantiating

-- 
John Carter
Phone : (64)(3) 358 6639
Tait Electronics
PO Box 1645 Christchurch
New Zealand

-- 

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