Any ways to achieve this?
Jameson於 2015年1月7日星期三UTC-8下午7時32分59秒寫道： > > 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] > <javascript:>> 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? >>>>> >>>>
