4 ms·
> There's even a show_expr method that prints it as an S-expression. Yes. However, the 99.9999% of Julia code out there isn't in s-expressions. Every Julia pr
by flavio81 6y ago
> There's even a show_expr method that prints it as an S-expression.
Yes.
However, the 99.9999% of Julia code out there isn't in s-expressions. Every Julia programmer doesn't use s-expressions. They use Julia syntax.
Thus, the task of writing Macros in Julia is harder than doing the same thing in Lisp.
>So I'm seeing nothing that would be problematic if it had just gone with a Lisp syntax.
They should've gone with a Lisp syntax, IMO. Beginners, who haven't passed through the chore of learning C-like syntax, find s-expressions as easy as any other kind of syntax.
- didibus 6y agoI'm with you, I just thought prior commenter was making a claim that Julia couldn't have been designed with a Lisp syntax, so I thought pointing out that it can already support representing all of its code as an s-expression kind of shows it could have very well had a Lisp syntax. I wonder if it's too late :p, maybe it can do what Racket does, and build on some per file #lang macro, give us proper Lisp syntax.