cannot deduce template lambda from argument

[aliak]

On Wednesday, 6 December 2017 at 11:02:01 UTC, Jonathan M Davis 
wrote:
> If you only want one type, then given n that type; I'm pretty 
> sure that it would be
>
> alias lambda = (int n) => n * n;
>
> if you wanted an int. But if you want to do anything more 
> complicated with it, it would make more sense to just turn it 
> into a proper function template.
>
> - Jonathan M Davis

Roight, I was more thinking along the lines of a little more 
complicated I guess :)

i.e.:

template lambda(T)
     if (isIntegral!T)
{
     alias lambda = (T n) => n * n;
}

But you're right, if it gets more complicated a proper function 
is probably better. Plus I just tried and it seems like you can't 
really constrain a variable template anyway. Unless there's some 
magic syntax I've overlooked in the docs.

Thanks!

Thanks!