Closed ldemailly closed 2 months ago
x => x*2
func (x) {x*2}
Fixes #135
I may bite the bullet and make
(x,y,z) => x+y+z
work but that's a big change to the parser
One minor remark, but seems OK
you meant awesome :)
thanks for the reviews!
x => x*2
==func (x) {x*2}
)Fixes #135