the declaration `const A = (2,3)` is not part of the macro environment. values are only seen by functions. whereas macros only see the surface syntax. so the arguments to `@f` are the literals `1` and `:(A...)`
On Wed Jan 07 2015 at 10:04:26 PM Chi-wei Wang <[email protected]> wrote: > Can ... be applied on macro? I tried this but it didn't compile. > macro f(a,b,c) > return :($a+$b+$c) > end > const A = (2,3) > @f(1, A...) > > > > Chi-wei Wang於 2015年1月7日星期三UTC-8下午6時45分14秒寫道： > >> It works! I missed the varargs section in the doc. Thanks! >> >> Greg Plowman於 2015年1月7日星期三UTC-8下午5時26分02秒寫道： >>> >>> >>> Would the following work? >>> >>> const A = (1,2,3) >>> f(x::Int, y::Int, z::Int) = x+y+z >>> f(A...) >>> >>> >>> http://docs.julialang.org/en/release-0.3/manual/functions/# >>> varargs-functions >>> >>> >>> On Thursday, January 8, 2015 11:14:23 AM UTC+11, Chi-wei Wang wrote: >>> >>>> Hi, everyone. I am trying to achieve the effect like the following C >>>> code: >>>> >>>> #define A 1,2,3 >>>> void f(int x, int y, int z) { >>>> } >>>> >>>> f(A); >>>> >>>> Yet the macro in Julia always returns a single expression. Is it >>>> possible to do this? >>>> >>>
