What makes macro different from function? Why is it not possible to do: foo(e::Expr) = :(&e) @foo x = x + 5
On Friday, June 3, 2016 at 10:05:46 AM UTC+2, Ford O. wrote: > > I think this deserves an own topic. > > *You* should post here syntax that looks like duplicate or you think it > could use already existing keyword. (mark with* # identical *or *# > replacement*) > Rule of thumb - *the less special syntax the better*. > > # identical > # replace ' , ' with ' ; ' in arrays ? > [1, 2, 3, 4] > [1; 2; 3; 4] > > > > # identical > # replace ' = ' with ' in ' in for loops ? > for i = 1:10 > for i in 1:10 > > > > > # replacement > # replace ' -> ' with ' = ' in anonymous functions ? > (a, b, c) -> a + b + c > (a, b, c) = a + b + c >
