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? >> >
