isInputRange copied verbatim produces a different result than isInputRange from std.range

[ag0aep6g]

On 03/04/2018 08:54 PM, aliak wrote:
> wait a minute... so I can't use any std.range functions on a type if I 
> add the range primitives as free functions? O.o

Yes. In other words: You can't implement range primitives as free 
functions. Because std.range (and std.algorithm, etc.) doesn't know 
about them.