You might want to file this as an issue. https://github.com/JuliaLang/julia/issues
--Tim On Saturday, September 13, 2014 05:10:08 AM Jonas Hagen wrote: > Ok, it should be: > > macro myDef3(name) > return esc(:( $name(x; y=3) = return (x*y) )) > end > @myDef3 f3 > f3(4) # now gives 12, as expected > f3(4, y=5) #now gives 20, as expected > > The return must be explicitly written. But why? I find this very confusing. > > On Monday, September 8, 2014 9:58:24 PM UTC+2, Jonas Hagen wrote: > > Hello! > > > > I'm trying to define a function using a macro. But I just dont get it. > > Look at the following Examples: > > > > Using default arguments, everything works as expected: > > macro myDef2(name) > > > > return esc(:( $name(x, y=3) = x*y )) > > > > end > > @myDef2 f2 > > f2(4) # gives 12, as expected > > f2(4, 5) #gives 20, as expected > > > > So, I assumed the same would work for keyword arguments, but it doesn't: > > macro myDef3(name) > > > > return esc(:( $name(x; y=3) = x*y )) > > > > end > > @myDef3 f3 > > f3(4) # gives 3 but 12 expected > > f3(4, y=5) #gives 5 but 20 expected > > > > Even more strange: > > macro myDef4(name) > > > > return esc(:( $name(x; y=3) = x )) # not using y here! > > > > end > > @myDef4 f4 # gives ERROR: syntax: malformed expression > > > > Am I doing something wrong or is this a bug? > > Maybe I just don't understand what esc() does or how to define functions > > in macros, I'm new to Julia. > > I'm using Version 0.3.0 (2014-08-20 20:43 UTC). > > > > - Jonas
