Hi,

I originally interpreted Erik's reply to mean -> Instead of `#(last (sort 
%&))` do `#((comp last sort) %&)`. This works without any issue. 

Herwig's suggestion was about replacing the entire anonymous function with 
`(comp last sort vec)`. The example you gave `((comp last sort) 3 2 1)`, 
even if we make it `((comp last sort vec) 3 2 1)`, it will throw an error 
about `vec` being called with too many parameters. 

For it to work with variable args, as Herwig said in his reply to me, we 
need a function that can return a sequence from a varargs. `vector` or 
`list would work, not `vec`. It should be `((comp last sort list) 3 2 1)`. 
Please correct me if I am understanding this wrong.



On Thursday, 13 August 2015 16:13:24 UTC+5:30, Tassilo Horn wrote:
>
> Amith George <strid...@gmail.com <javascript:>> writes: 
>
> >> That's not the same function as #(last (sort %&)) 
> >> The equivalent would be (comp last sort vec) 
> > 
> > Could you please explain why is the `vec` needed?  From what I 
> > understand, we are expected to treat the variadic args argument as a 
> > seq, nothing more. 
>
> With ((comp last sort) 3 2 1), sort will be called as 
>
>   (apply sort (list 3 2 1)) 
>
> which essentially is 
>
>   (sort 3 2 1). 
>
> Hovever, sort is no varargs function but wants a collection. 
>
> Bye, 
> Tassilo 
>

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