Thread overview
alias with lambda syntax: alias fun2=a=>fun(a);
Jun 05, 2014
Timothee Cour
Jun 05, 2014
timotheecour
Jun 06, 2014
Meta
June 05, 2014
Is there a way to do this?

import std.algorithm;

auto fun(T)(T a){return a;}

template fun2(T){auto fun2(T a){return fun(a);}}//OK but heavy syntax and
cannot be nested inside test()

void main(){
  //alias fun2=fun!int; //OK but needs to specify template params
  //none of those work:
  //alias fun2=a=>fun(a);
  //alias fun2(T)=(T a)=>fun(a);
  //alias fun2(T)=(T a){return fun(a);}
  auto b=[1].map!fun2;
  assert(b.equal([1]));
}


June 05, 2014
On Thursday, 5 June 2014 at 06:58:50 UTC, Timothee Cour via
Digitalmars-d-learn wrote:
> Is there a way to do this?
>
> import std.algorithm;
>
> auto fun(T)(T a){return a;}
>
> template fun2(T){auto fun2(T a){return fun(a);}}//OK but heavy syntax and
> cannot be nested inside test()
>
> void main(){
>   //alias fun2=fun!int; //OK but needs to specify template params
>   //none of those work:
>   //alias fun2=a=>fun(a);
>   //alias fun2(T)=(T a)=>fun(a);
>   //alias fun2(T)=(T a){return fun(a);}
>   auto b=[1].map!fun2;
>   assert(b.equal([1]));
> }

ok I remembered we can use std.typetuple.Alias for that.
June 06, 2014
On Thursday, 5 June 2014 at 07:19:07 UTC, timotheecour wrote:
> ok I remembered we can use std.typetuple.Alias for that.

Or std.functional.unaryFun.