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 t
On Wednesday, December 06, 2017 10:43:18 aliak via Digitalmars-d-learn
wrote:
> On Wednesday, 6 December 2017 at 08:10:26 UTC, Biotronic wrote:
> > On Tuesday, 5 December 2017 at 23:01:43 UTC, aliak wrote:
> >> immutable lambda(T) = (T n) => n * n;
> >
> > Generally, you'd want to write
> >
> >
On Wednesday, 6 December 2017 at 08:10:26 UTC, Biotronic wrote:
On Tuesday, 5 December 2017 at 23:01:43 UTC, aliak wrote:
immutable lambda(T) = (T n) => n * n;
Generally, you'd want to write
alias lambda = n => n * n;
instead. That said, I don't see any reason why your syntax
shouldn't
On Tuesday, 5 December 2017 at 23:01:43 UTC, aliak wrote:
immutable lambda(T) = (T n) => n * n;
Generally, you'd want to write
alias lambda = n => n * n;
instead. That said, I don't see any reason why your syntax
shouldn't work, and would argue it's a bug. Please file it in
Bugzilla.
Hi,
Having a little trouble understanding lambda type deduction. I
have this lambda:
immutable lambda(T) = (T n) => n * n;
and if I call it with an explicit type it works else it errors
with: lambda cannot deduce function from argument types !()(int)
auto x = lambda!int(2); // ok
auto x =
